IB Math Analysis & Approaches Question bank -Topic: SL 1.7 Laws of exponents- SL Paper 1

Question

Find \({\log _2}32\) .

[1]
a.

Given that \({\log _2}\left( {\frac{{{{32}^x}}}{{{8^y}}}} \right)\) can be written as \(px + qy\) , find the value of p and of q.

[4]
b.
Answer/Explanation

Markscheme

5     A1     N1

[1 mark]

a.

METHOD 1

\({\log _2}\left( {\frac{{{{32}^x}}}{{{8^y}}}} \right) = {\log _2}{32^x} – {\log _2}{8^y}\)     (A1)

\( = x{\log _2}32 – y{\log _2}8\)     (A1)

\({\log _2}8 = 3\)     (A1)

\(p = 5\) , \(q = – 3\) (accept \(5x – 3y\) )     A1      N3 

METHOD 2

\(\frac{{{{32}^x}}}{{{8^y}}} = \frac{{{{({2^5})}^x}}}{{{{({2^3})}^y}}}\)     (A1) 

\( = \frac{{{2^5}^x}}{{{2^3}^y}}\)     (A1)

\( = {2^{5x – 3y}}\)     (A1)

\({\log _2}({2^{5x – 3y}}) = 5x – 3y\)

\(p = 5\) , \(q = – 3\) (accept \(5x – 3y\) )     A1      N3

[4 marks]

b.

Question

Solve \({\log _2}x + {\log _2}(x – 2) = 3\) , for \(x > 2\) .

Answer/Explanation

Markscheme

recognizing \(\log a + \log b = \log ab\) (seen anywhere)     (A1)

e.g. \({\log _2}(x(x – 2))\) , \({x^2} – 2x\)

recognizing \({\log _a}b = x \Leftrightarrow {a^x} = b\)     (A1)

e.g. \({2^3} = 8\)

correct simplification     A1

e.g. \(x(x – 2) = {2^3}\) , \({x^2} – 2x – 8\)

evidence of correct approach to solve     (M1)

e.g. factorizing, quadratic formula

correct working     A1

e.g. \((x – 4)(x + 2)\) , \(\frac{{2 \pm \sqrt {36} }}{2}\)

\(x = 4\)     A2     N3

[7 marks]

 

Question

Write down the value of

(i)     \({\log _3}27\);

[1]
a(i).

(ii)     \({\log _8}\frac{1}{8}\);

[1]
a(ii).

(iii)     \({\log _{16}}4\).

[1]
a(iii).

Hence, solve \({\log _3}27 + {\log _8}\frac{1}{8} – {\log _{16}}4 = {\log _4}x\).

[3]
b.
Answer/Explanation

Markscheme

(i)     \({\log _3}27 = 3\)     A1     N1

[1 mark]

a(i).

(ii)     \({\log _8}\frac{1}{8} =  – 1\)     A1     N1

[1 mark]

a(ii).

(iii)     \({\log _{16}}4 = \frac{1}{2}\)     A1     N1

[1 mark]

a(iii).

correct equation with their three values     (A1)

eg     \(\frac{3}{2} = {\log _4}x{\text{, }}3 + ( – 1) – \frac{1}{2} = {\log _4}x\)

correct working involving powers     (A1)

eg     \(x = {4^{\frac{3}{2}}}{\text{, }}{4^{\frac{3}{2}}} = {4^{{{\log }_4}x}}\)

\(x = 8\)     A1     N2

[3 marks]

b.

Question

Find the value of each of the following, giving your answer as an integer.

\({\log _6}36\)

[2]
a.

\({\log _6}4 + {\log _6}9\)

[2]
b.

\({\log _6}2 – {\log _6}12\)

[3]
c.
Answer/Explanation

Markscheme

correct approach     (A1)

eg     \({6^x} = 36,{\text{ }}{6^2}\)

\(2\)      A1     N2

[2 marks]

a.

correct simplification     (A1)

eg     \({\log _6}36,{\text{ }}\log (4 \times 9)\)

\(2\)      A1      N2

[2 marks]

b.

correct simplification     (A1)

eg     \({\log _6}\frac{2}{{12}},{\text{ }}\log (2 \div 12)\)

correct working     (A1)

eg     \({\log _6}\frac{1}{6},{\text{ }}{6^{ – 1}} = \frac{1}{6}{,6^x} = \frac{1}{6}\)

\(-1\)     A1     N2

[3 marks]

c.

Question

Let \(x = \ln 3\) and \(y = \ln 5\). Write the following expressions in terms of \(x\) and \(y\).

\(\ln \left( {\frac{5}{3}} \right)\).

[2]
a.

\(\ln 45\).

[4]
b.
Answer/Explanation

Markscheme

correct approach     (A1)

eg\(\,\,\,\,\,\)\(\ln 5 – \ln 3\)

\(\ln \left( {\frac{5}{3}} \right) = y – x\)    A1     N2

[2 marks]

a.

recognizing factors of 45 (may be seen in log expansion)     (M1)

eg\(\,\,\,\,\,\)\(\ln (9 \times 5),{\text{ }}3 \times 3 \times 5,{\text{ }}\log {3^2} \times \log 5\)

correct application of \(\log (ab) = \log a + \log b\)     (A1)

eg\(\,\,\,\,\,\)\(\ln 9 + \ln 5,{\text{ }}\ln 3 + \ln 3 + \ln 5,{\text{ }}\ln {3^2} + \ln 5\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(2\ln 3 + \ln 5,{\text{ }}x + x + y\)

\(\ln 45 = 2x + y\)    A1     N3

[4 marks]

b.

Question

An arithmetic sequence has \({u_1} = {\text{lo}}{{\text{g}}_c}\left( p \right)\) and \({u_2} = {\text{lo}}{{\text{g}}_c}\left( {pq} \right)\), where \(c > 1\) and \(p,\,\,q > 0\).

Show that \(d = {\text{lo}}{{\text{g}}_c}\left( q \right)\).

[2]
a.

Let \(p = {c^2}\) and \(q = {c^3}\). Find the value of \(\sum\limits_{n = 1}^{20} {{u_n}} \).

[6]
b.
Answer/Explanation

Markscheme

valid approach involving addition or subtraction       M1
eg  \({u_2} = {\text{lo}}{{\text{g}}_c}\,p + d,\,\,{u_1} – {u_2}\)

correct application of log law      A1
eg  \({\text{lo}}{{\text{g}}_c}\left( {pq} \right) = {\text{lo}}{{\text{g}}_c}\,p + {\text{lo}}{{\text{g}}_c}\,q,\,\,{\text{lo}}{{\text{g}}_c}\left( {\frac{{pq}}{p}} \right)\)

\(d = {\text{lo}}{{\text{g}}_c}\,q\)    AG N0

[2 marks]

a.

METHOD 1 (finding \({u_1}\) and d)

recognizing \(\sum { = {S_{20}}} \) (seen anywhere)      (A1)

attempt to find \({u_1}\) or d using \({\text{lo}}{{\text{g}}_c}\,{c^k} = k\)     (M1)
eg  \({\text{lo}}{{\text{g}}_c}\,c\), \({\text{3}}\,{\text{lo}}{{\text{g}}_c}\,c\), correct value of \({u_1}\) or d

\({u_1}\) = 2, d = 3 (seen anywhere)      (A1)(A1)

correct working     (A1)
eg  \({S_{20}} = \frac{{20}}{2}\left( {2 \times 2 + 19 \times 3} \right),\,\,{S_{20}} = \frac{{20}}{2}\left( {2 + 59} \right),\,\,10\left( {61} \right)\)

\(\sum\limits_{n = 1}^{20} {{u_n}} \) = 610     A1 N2

METHOD 2 (expressing S in terms of c)

recognizing \(\sum { = {S_{20}}} \) (seen anywhere)      (A1)

correct expression for S in terms of c      (A1)
eg  \(10\left( {2\,{\text{lo}}{{\text{g}}_c}\,{c^2} + 19\,{\text{lo}}{{\text{g}}_c}\,{c^3}} \right)\)

\({\text{lo}}{{\text{g}}_c}\,{c^2} = 2,\,\,\,{\text{lo}}{{\text{g}}_c}\,{c^3} = 3\)  (seen anywhere)     (A1)(A1)

correct working      (A1)

eg  \({S_{20}} = \frac{{20}}{2}\left( {2 \times 2 + 19 \times 3} \right),\,\,{S_{20}} = \frac{{20}}{2}\left( {2 + 59} \right),\,\,10\left( {61} \right)\)

\(\sum\limits_{n = 1}^{20} {{u_n}} \) = 610     A1 N2

METHOD 3 (expressing S in terms of c)

recognizing \(\sum { = {S_{20}}} \) (seen anywhere)      (A1)

correct expression for S in terms of c      (A1)
eg  \(10\left( {2\,{\text{lo}}{{\text{g}}_c}\,{c^2} + 19\,{\text{lo}}{{\text{g}}_c}\,{c^3}} \right)\)

correct application of log law     (A1)
eg  \(2\,{\text{lo}}{{\text{g}}_c}\,{c^2} = \,\,{\text{lo}}{{\text{g}}_c}\,{c^4},\,\,19\,{\text{lo}}{{\text{g}}_c}\,{c^3} = \,\,{\text{lo}}{{\text{g}}_c}\,{c^{57}},\,\,10\,\left( {{\text{lo}}{{\text{g}}_c}\,{{\left( {{c^2}} \right)}^2} + \,\,{\text{lo}}{{\text{g}}_c}\,{{\left( {{c^3}} \right)}^{19}}} \right),\,\,10\,\left( {{\text{lo}}{{\text{g}}_c}\,{c^4} + \,{\text{lo}}{{\text{g}}_c}\,{c^{57}}} \right),\,\,10\left( {{\text{lo}}{{\text{g}}_c}\,{c^{61}}} \right)\)

correct application of definition of log      (A1)
eg  \({\text{lo}}{{\text{g}}_c}\,{c^{61}} = 61,\,\,{\text{lo}}{{\text{g}}_c}\,{c^4} = 4,\,\,{\text{lo}}{{\text{g}}_c}\,{c^{57}} = 57\)

correct working     (A1)
eg  \({S_{20}} = \frac{{20}}{2}\left( {4 + 57} \right),\,\,10\left( {61} \right)\)

\(\sum\limits_{n = 1}^{20} {{u_n}} \) = 610     A1 N2

[6 marks]

b.

MAA SL 1.7 BINOMIAL THEOREM [concise]-lala

Question

[Maximum mark: 9] [without GDC]
Write down the expansions of

(a) \((1\pm x)^{3}\)

(b)  \((1\pm x)^{4}\)

(c)  \((1\pm x)^{5}\)

Answer/Explanation

Ans.

(a) \((1\pm x)^{3}=1\pm 3x+3x^{2}\pm x^{3}\)

(b) \((1\pm x)^{4}=1\pm 4x+6x^{2}\pm 4x^{3}+x^{4}\)

(c) \((1\pm x)^{5}=1\pm 5x+10x^{2}\pm 10x^{3}+5x^{4}\pm x^{5}\)

Question

[Maximum mark: 10] [with GDC]
Write down the first four terms in ascending powers of x

(a) in the expansion of \((1+x)^{10}\)

(b) in the expansion of \((1+2x)^{10}\)

(c) in the expansion of \((2-x)^{10}\)

Answer/Explanation

Ans.

(a) \((1+x)^{10}=1+10x+45x^{2}+120x^{3}+…\)

(b) \((1+2x)^{10}=1+20x+180x^{2}+960x^{3}+…\)

(c) \((2-x)^{10}=1024-5120x+11520x^{2}-15360x^{3}+…\)

Question

[Maximum mark: 5] [with GDC]

Find the term in  x7 in the expansion of \((1-x)^{10}\)

Answer/Explanation

Ans.

-120x7

Question

[Maximum mark: 5] [with GDC]

Find the term in  x16 in the expansion of \((1-x^{2})^{10}\)

Answer/Explanation

Ans.

45x16

Question

[Maximum mark: 10] [with GDC]

(a) Expand      (i) \(\left ( x-\frac{1}{x} \right )^{3}\)                    (ii) \(\left ( x-\frac{1}{x} \right )^{4}\)

(b) In the expansion of  \(\left ( x-\frac{1}{x} \right )^{10}\)

(i) find the constant term
(ii) find the term in x2
(iii) find the term in x-2

Answer/Explanation

Ans.

(a) (i) \(\left ( x-\frac{1}{x} \right )^{3}=x^{3}-3x+\frac{3}{x}-\frac{1}{x^{3}}\)                (ii)  \(\left ( x-\frac{1}{x} \right )^{4}=x^{4}-4x^{2}+6-\frac{4}{x^{2}}-\frac{1}{x^{4}}\)

(b) (i) \(-\binom{10}{5}=-252\)           (ii) \(\binom{10}{4}=210\)             (iii) \(\binom{10}{6}=210\)

 

Question

[Maximum mark: 10] [with GDC]

(a) Expand  \((2x+1)^{4}\)

Hence, find
(b) the term in x2 in the expansion of \(x(2x+1)^{4}\)

(c) the term in x2 in the expansion of \((x+1)(2x+1)^{4}\)

(d) the term in x2 in the expansion of \((3-x^{2})(2x+1)^{4}\)

Answer/Explanation

Ans.

(a) \((1+2x)^{4}=1+8x+24x^{2}+32x^{3}+16x^{4}\)                (b)  8x2             (c) 32x2            (d) 71x2

Question

[Maximum mark: 10] [without GDC]
(a) Show that
(i) \(\binom{8}{7}=8\),                     (ii) \(\binom{8}{6}=28\),                      (iii) \(\binom{8}{5}=56\)
(b) Find
(i) \(\binom{10}{9}\),                                      (ii) \(\binom{10}{8}\),                                            (iii) \(\binom{10}{7}\)

Answer/Explanation

Ans.

(a) (show)
(b)        (i) 10          (ii) 45           (iii) 120

Question

[Maximum mark: 8] [without GDC]

(a) Show that

(i) \(\binom{n}{0}=1\),                        (ii) \(\binom{n}{1}=n\),                      (iii) \(\binom{n}{2}=\frac{n(n-1)}{2}\)

(b) Simplify \(\binom{n}{3}\)

Answer/Explanation

Ans.

(a) (show)

(b) \(\binom{n}{3}=\frac{n(n-1)(n-2)}{6}\)

Question

[Maximum mark: 8] [without GDC]

(a) Verify that

\(\binom{5}{2}+\binom{5}{3}=\binom{6}{3}\)

(b) Prove that

\(\binom{19}{9}+\binom{19}{10}=\binom{20}{10}\)

Answer/Explanation

Ans.

 

(a) (verify)

(b) LHS= \(\binom{19}{9}+\binom{19}{10}=\frac{19!}{9!x10!}+\frac{19!}{9!x10!}=\frac{2×19!}{9!x10!}=\frac{2×19!}{9!x10!}\)

RHS= \(\binom{20}{10}=\frac{20!}{10!x10!}=\frac{20×19!}{10×9!x10!}=\frac{2×19!}{9!x10!}\)

Question

[Maximum mark: 6] [with / without GDC]

Complete the following expansion.

\((2+ax)^{4}=16+32ax+…\)

Answer/Explanation

Ans.

\(…+6×2^{2}(ax)^{2}+4×2(ax)^{3}+(ax)^{4}=…+24a^{2}x^{2}+8a^{3}x^{3}+a^{4}x^{4}\)

Question

[Maximum mark: 4] [with GDC]

Use the binomial theorem to complete this expansion.

\((3x+2y)^{4}=81x^{4}+216x^{3}y+…\)

Answer/Explanation

Ans.

\((3x+2y)^{4}=(3x)^{4}+\binom{4}{1}(3x)^{2}(2y)+\binom{4}{2}(3x)^{2}(2y)^{2}+\binom{4}{3}(3x)(2y)^{3}+(2y)^{4}\)
=\(81x^{4}+216x^{3}y+216x^{2}y^{2}+96xy^{3}+16y^{4}\)

Question

[Maximum mark: 6] [with / without GDC]

Given that \((3+\sqrt{7})^{3}=p+q\sqrt{7}\) where \(p\) and \(q\) are integers, find

(a) \(p\);                         (b) \(q\).

Answer/Explanation

Ans.

METHOD 1

Using binomial expansion

\((3+\sqrt{7})^{3}=3^{3}+\binom{3}{1}3^{2}(\sqrt{7})+\binom{3}{2}3(\sqrt{7})^{2}+(\sqrt{7})^{3}=27+27\sqrt{7}+63+7\sqrt{7}\)

\((3+\sqrt{7})^{3}=90+34\sqrt{7}\)                   (so p=90, q=34)

 

METHOD 2

For multiplying

\((3+\sqrt{7})^{2}(3+\sqrt{7})=(9+6\sqrt{7}+7)(3+\sqrt{7})=27+9\sqrt{7}+18\sqrt{7}+42+21+7\sqrt{7}\)

\((3+\sqrt{7})^{3}=90+34\sqrt{7}\)                  (so p=90, q=34)

Question

[Maximum mark: 4] [with / without GDC]

Find the coefficient of a5b7 in the expansion of \((a+b)^{12}\).

Answer/Explanation

Ans.

(a + b)12

Coefficient of a5b7 is \(\binom{12}{5}=\binom{12}{7}=792\)

Question

[Maximum mark: 4] [with / without GDC]

Find the coefficient of a3b4 in the expansion of \((5a+b)^{7}\).

Answer/Explanation

Ans.

\((5a+b)^{7}=…+\binom{7}{4}(5a)^{3}(b)^{4}+…=\frac{7x6x5x4}{1x2x3x4}x5^{3}x(a^{3}b^{4})=35×5^{3}xa^{3}b^{4}\)

So the coefficient is 4375

Question

[Maximum mark: 6] [without GDC]

The fifth term in the expansion of the binomial \((a+b)^{n}\) is given by \(\binom{10}{4}p^{6}(2q)^{4}\).

(a) Write down the value of \(n\).
(b) Write down \(a\) and \(b\), in terms of \(p\) and/or \(q\).
(c) Write down an expression for the sixth term in the expansion.

Answer/Explanation

Ans.

(a) n = 10
(b) a = p, b = 2q (or a = 2q, b = p)

(c) \(\binom{10}{5}p^{5}(2q)^{5}\)

Question

[Maximum mark: 5] [with / without GDC]

Consider the expansion of \((x+2)^{11}.\)

(a) Write down the number of terms in this expansion.
[b) Find the term containing x2.

Answer/Explanation

Ans.

(a) 12 terms

(b) (10th term, \(r=9, \binom{11}{9}(x)^{2}(2)^{9})\)
\(\binom{11}{9}(x)^{2}(2)^{9}=55×2^{9}x^{2}=28160x^{2}\)

Question

[Maximum mark: 6] [with / without GDC]

Find the coefficient of x3 in the expansion of \((2-x)^{5}\).

Answer/Explanation

Ans.

Term involving x3 is  \(\binom{5}{3}(2)^{2}(-x)^{3}=-40x^{3}\Rightarrow\) The coefficient is –40

Question

[Maximum mark: 4] [with GDC]

Find the coefficient of x5 in the expansion of \((3x-2)^{8}\).

Answer/Explanation

Ans.

Required term is \(\binom{8}{5}(3x)^{5}(-2)^{3}\)

Therefore the coefficient of x5 is 56 × 243 × –8 = –108864

Question

[Maximum mark: 6] [with GDC]

Find the term containing x3 in the expansion of \((2-3x)^{8}\).

Answer/Explanation

Ans.

\(\binom{8}{3}(2)^{5}(-3x)^{3}\)
Term is \(-48384x^{3}\)

Question

[Maximum mark: 5] [with GDC]

Find the term in x3 in the expansion of \(\left ( \frac{2}{3}x-3 \right )^{8}\).

Answer/Explanation

Ans.

\(\binom{8}{5}\left ( \frac{2}{3}x \right )^{3}(-3)^{5}=-4032x^{3}\)

Question

[Maximum mark: 6] [with / without GDC]

One of the terms of the expansion of \((x+2y)^{10}\) is \(ax^{8}y^{2}\). Find the value of \(a\).

Answer/Explanation

Ans.

\(a=\binom{10}{8}x2^{2}=45×4=180\)

Question

[Maximum mark: 6] [with GDC]

Find the term containing x10 in the expansion of \((5+2x^{2})^{7}\).

Answer/Explanation

Ans.

\(\left ( \frac{7}{2} \right )5^{2}(2x^{2})^{5}=16\;800x^{10}\)

Question

[Maximum mark: 6] [with GDC]

Consider the expansion of  \((x^{2}-2)^{5}\).

(a) Write down the number of terms in this expansion.
(b) The first four terms of the expansion in descending powers of \(x\) are

\(x^{10}-10x^{8}+40x^{6}+Ax^{4}+…\)

Find the value of \(A\).

Answer/Explanation

Ans.

(a) 6 terms

(b) the fourth term is \(\binom{5}{3}(-2)^{3}(x^{2})^{2}=10(-8)x^{4}=-80x^{4}\)       hence \(A\)=-80

Question

[Maximum mark: 6] [with GDC]

Consider the expansion of \(\left ( 3x^{2}-\frac{1}{x} \right )^{9}\).

(a) How many terms are there in this expansion?
(b) Find the constant term in this expansion.

Answer/Explanation

Ans.

(a) 10

(b)  \(\binom{9}{6}(3x^{2})^{3}\left ( -\frac{1}{x} \right )^{6}=84×3^{3}x^{6}\frac{1}{x^{6}}\)

constant =2268

Question

[Maximum mark: 4] [with GDC]

Determine the constant term in the expansion of \(\left ( x-\frac{2}{x^{2}} \right )^{9}\).

Answer/Explanation

Ans.

\(\binom{9}{3}x^{6}\left ( \frac{-2}{x^{2}} \right )^{3}=84x^{6}\left ( \frac{-8}{x^{6}} \right )=-672\)

Question

[Maximum mark: 6] [with GDC]

Consider the expansion of the expression \((x^{3}-3x)^{6}\).

(a) Write down the number of terms in this expansion.
(b) Find the term in x12.

Answer/Explanation

Ans.

(a) 7 terms

(b) \(\binom{6}{3}(x^{3})^{3}(-3x)^{3}\)
\(\binom{6}{3}=20,(-3)^{3}=-27\)

Term is -540x12

Question

[Maximum mark: 6] [with / without GDC]

Find the term in x4 in the expansion of \(\left ( 3x^{2}-\frac{2}{x} \right )^{5}\).

Answer/Explanation

Ans.

\(\binom{5}{2}(3x^{2})^{3}\left ( \frac{-2}{x} \right )^{2}\binom{5}{2},27x^{6},\frac{4}{x^{2}};10(3x^{2})^{3}\left ( \frac{-2}{x} \right )^{2}=10x27x4x^{4}\)

term = 1080x4

Question

[Maximum mark: 6] [with GDC]

When the expression \((2+ax)^{10}\) is expanded, the coefficient of the term in x3 is 414 720.
Find the value of \(a\).

Answer/Explanation

Ans.

\(\binom{10}{3}2^{7}(ax)^{3}\)                  \( \left ( accept\binom{10}{7} \right )\)
\(120×2^{7}a^{3}=414\;720\)
\(a^{3}=27\Leftrightarrow a=3\)

Question

[Maximum mark: 6] [without GDC]

(a) Expand \((x-2)^{4}\) and simplify your result.

(b) Find the term in x3 in \((3x+4)(x-2)^{4}\).

Answer/Explanation

Ans.

(a) \((x-2)^{4}=x^{4}+4x^{3}(-2)+6x^{2}(-2)^{2}+4x(-2)^{3}+(-2)^{4}\)
\((x-2)^{4}=x^{4}-8x^{3}+24x^{2}-32x+16\)

(b) finding coefficients, 3 × 24 (= 72), 4 × (–8)(= –32)
term is 40x3

Question

[Maximum mark: 6] [without GDC]

(a) Expand \((2+x)^{4}\) and simplify your result.

(b) Hence, find the term in x2 in \((2+x)^{4}\left ( 1+\frac{1}{x^{2}} \right )\).

Answer/Explanation

Ans.

(a)  \(2^{4}+4(2^{3})x+6(2^{2})x^{2}+4(2)x^{3}+x^{4},(4+4x+x^{2})(4+4x+x^{2})\)
\((2+x)^{4}=16+32x+24x^{2}+8x^{3}+x^{4}\)

(b) finding coefficients 24 and 1
term is 25x2

A

Question

[Maximum mark: [4] [without GDC]

Consider the binomial expansion \((1+x)^{4}=1+\binom{4}{1}x+\binom{4}{2}x^{2}+\binom{4}{3}x^{3}+x^{4}\)

(a) By substituting x = 1 into both sides, or otherwise, evaluate \(\binom{4}{1}+\binom{4}{2}+\binom{4}{3}\).

(b) Evaluate \(\binom{9}{1}+\binom{9}{2}+\binom{9}{3}+\binom{9}{4}+\binom{9}{5}+\binom{9}{6}+\binom{9}{7}+\binom{9}{8}\).

Extra question

Find a similar result by substituting x =- 1 in the expansion of (1+ x)4 above.

Answer/Explanation

Ans.

(a)  \((1+1)^{4}=2^{4}=1+\binom{4}{1}(1)+\binom{4}{2}(1^{2})+\binom{4}{3}(1^{3})+1^{4}\)
\(\Rightarrow \binom{4}{1}+\binom{4}{2}+\binom{4}{3}=16-2=14\)

 

(b)   \((1+1)^{9}=1+\binom{9}{1}+\binom{9}{2}+\binom{9}{3}+…+\binom{9}{8}+1\)
\(\Rightarrow \binom{9}{1}+\binom{9}{2}+\binom{9}{3}+…+\binom{9}{8}=2^{9}-2=510\)

Extra question

\(0=1-\binom{4}{1}+\binom{4}{2}-\binom{4}{3}+1\Rightarrow \binom{4}{1}-\binom{4}{2}+\binom{4}{3}=2\)

Question

[Maximum mark: 6] [without GDC]

(a) Expand \(\left ( e+\frac{1}{e} \right )^{4}\) in terms of e.

(b) Express \(\left ( e+\frac{1}{e} \right )^{4}+\left ( e-\frac{1}{e} \right )^{4}\) as the sum of three terms.

Answer/Explanation

Ans.

(a)  \(\left ( e+\frac{1}{e} \right )^{4}=\binom{4}{0}e^{4}+\binom{4}{1}e^{3}\left ( \frac{1}{e} \right )+\binom{4}{2}e^{2}\left ( \frac{1}{e} \right )^{2}+\binom{4}{3}e\left ( \frac{1}{e} \right )^{3}+\binom{4}{4}\left ( \frac{1}{e} \right )^{4}\)

\(\left ( e+\frac{1}{e} \right )^{4}=e^{4}+4e^{3}\left ( \frac{1}{e} \right )+6e^{2}\left ( \frac{1}{e} \right )^{2}+4e\left ( \frac{1}{e} \right )^{3}+\left ( \frac{1}{e} \right )^{4}=e^{4}+4e^{2}+6+\frac{4}{e^{2}}+\frac{1}{e^{4}}\)

(b) \(\left ( e-\frac{1}{e} \right )^{4}=e^{4}-4e^{3}\left ( \frac{1}{e} \right )+6e^{2}\left ( \frac{1}{e} \right )^{2}-4e\left ( \frac{1}{e} \right )^{3}+\left ( \frac{1}{e} \right )^{4}=e^{4}-4e^{2}+6-\frac{4}{e^{2}}+\frac{1}{e^{4}}\)

Adding gives \(2e^{4}+12+\frac{2}{e^{4}}\)

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