IIT JEE Main Maths -Unit 5 - Binomial Theorem and Its Simple Applications- Exam Style Questions- New Syllabus

Question

If $ \sum_{r=0}^{11} \frac{C_{2r+1}}{2^{2r}} = \frac{m}{n} $, $ \gcd(m, n) = 1 $, then $ m – n $ is equal to ______.

▶️ Answer/Explanation

Ans. (2035)

Sol.

$ \int_{0}^{1} \left( C_0 x + \frac{C_1 x^2}{2} + \frac{C_2 x^3}{2^2} + \dots + \frac{C_{11} x^{12}}{2^{11}} \right) (1 + x) \, dx = \frac{C_0}{2} – \frac{C_1}{3} + \frac{C_2}{4} – \dots + \frac{C_{11}}{12} $

$ \int_{0}^{1} \left( C_0 x + \frac{C_1 x^2}{2} + \frac{C_2 x^3}{2^2} + \dots + \frac{C_{11} x^{12}}{2^{11}} \right) (1 – x) \, dx = C_0 – \frac{C_1}{2} + \frac{C_2}{3} – \dots + \frac{C_{11}}{12} $

$ \left( \frac{C_1}{2} + \frac{C_3}{2^2} + \frac{C_5}{2^3} + \dots + \frac{C_{11}}{2^6} \right) = \frac{\left( C_0 + \frac{C_2}{2} + \frac{C_4}{2^2} + \dots + \frac{C_{10}}{2^5} + \frac{C_{12}}{2^6} – \frac{C_1}{2} + \frac{C_3}{2^2} – \frac{C_5}{2^3} + \dots – \frac{C_{11}}{2^6} \right)}{2} $

$ = \frac{C_0 + \frac{C_2}{2} + \frac{C_4}{2^2} + \dots + \frac{C_{10}}{2^5} – \frac{C_1}{2} + \frac{C_3}{2^2} – \frac{C_5}{2^3} + \dots – \frac{C_{11}}{2^6}}{2} = \frac{2047}{12} $

$ m = 2047, n = 12 $

$ m – n = 2047 – 12 = 2035 $

Question

Let $\alpha$, $\beta$, $\gamma$ and $\delta$ be the coefficients of $x^7$, $x^5$, $x^3$ and $x$ respectively in the expansion of $(x + \sqrt{x} – 1)^5 (x – \sqrt{x} – 1)^5$, $x > 1$. If $u$ and $v$ satisfy the equations $\alpha u + \beta v = 18$, $\gamma u + \delta v = 20$, then $u + v$ equals :
(1) 5
(2) 4
(3) 3
(4) 8

▶ Answer/Explanation

Ans. (1)
Sol. $(x + \sqrt{x} – 1)^5 (x – \sqrt{x} – 1)^5$
= 2$\{^5C_0.x^5 + ^5C_2.x^3(x^3 – 1) + ^5C_4.x(x^3 – 1)^2\}$
= 2$\{5x^7 + 10x^6 + x^5 – 10x^4 – 10x^3 + 5x\}$
$\Rightarrow$ $\alpha = 10$, $\beta = 2$, $\gamma = -20$, $\delta = 10$
Now, $10u + 2v = 18$
$-20u + 10v = 20$
$\Rightarrow u = 1$, $v = 4$
$u + v = 5$

Question

If $\sum_{r=1}^{30} \frac{r^2 C_{30}}{C_{r-1}} = \alpha \times 2^{29}$, then $\alpha$ is equal to ______.

▶ Answer/Explanation

Ans. (465)

$ \sum_{r=1}^{30} \frac{r^2 \left( {}^{30}C_r \right)^2}{{}^{30}C_{r-1}} $

$ = \sum_{r=1}^{30} r^2 \frac{ \left( \frac{31 – r}{r} \right) 30! }{ r! (30 – r)! } $

$ \because \frac{{}^{30}C_r}{{}^{30}C_{r-1}} = \frac{30 – r + 1}{r} = \frac{31 – r}{r} $

$ = \sum_{r=1}^{30} (31 – r) \frac{30!}{(r – 1)! (30 – r)!} $

$ = 30 \sum_{r=1}^{30} \frac{(31 – r) 29!}{(r – 1)! (30 – r)!} $

$ = 30 \sum_{r=1}^{30} (30 – r + 1) \, {}^{29}C_{30 – r} $

$ = 30 \left( \sum_{r=1}^{30} (31 – r) \, {}^{29}C_{30 – r}
+ \sum_{r=1}^{30} {}^{29}C_{30 – r} \right) $

$ = 30 (29 \times 2^{28} + 2^{29}) = 30 (29 + 2) 2^{28} $

$ = 15 \times 31 \times 2^{29} $

$ = 465 (2^{29}) $

$ \alpha = 465 $

Question

The sum of all rational terms in the expansion of $(1 + 2^{1/3} + 3^{1/2})^6$ is equal to _______

▶️ Answer/Explanation

Ans. (612)

Sol.
$ (1 + 2^{1/3} + 3^{1/2})^6 $

$ = \sum \frac{6!}{r_1! \, r_2! \, r_3!} (1)^{r_1} (2^{1/3})^{r_2} (3^{1/2})^{r_3} $

$
\begin{array}{ccc}
r_1 & r_2 & r_3 \\
\hline
6 & 0 & 0 \\
4 & 0 & 2 \\
2 & 0 & 4 \\
0 & 0 & 6 \\
3 & 3 & 0 \\
1 & 3 & 2 \\
0 & 6 & 0 \\
\end{array}
$

$
\text{sum} =
\frac{6!}{6!0!0!}(1)^6(2)^0(3)^0 +
\frac{6!}{4!0!2!}(1)^4(2)^0(3)^2 +
\frac{6!}{2!0!4!}(1)^2(2)^0(3)^4 +
\frac{6!}{0!0!6!}(1)^0(2)^0(3)^6 +
\frac{6!}{3!3!0!}(1)^3(2)^1(3)^0 +
\frac{6!}{1!3!2!}(1)^1(2)^1(3)^2 +
\frac{6!}{0!6!0!}(1)^0(2)^2(3)^0
$

$ = 1 + 45 + 135 + 27 + 40 + 360 + 4 = 612 $

IIT JEE Main Maths -Unit 5 - Binomial Theorem and Its Simple Applications- Exam Style Questions- New Syllabus

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