Questions
Consider a geometric sequence with first term 1 and common ratio 10 .
Sn is the sum of the first n terms of the sequence.
(a)Find an expression for Sn in the form \(\frac{a^{n}-1}{b}\), where \(a,b\in \mathbb{Z^{+}}\).
(b) Hence, show that \(S_{1}+S_{2}+S_{3}+…+S_{n}=\frac{10(10^{n}-1)-9n}{81}\).
▶️Answer/Explanation
Detailed solution
Let’s dive into this geometric sequence problem with a sense of excitement! We’re given a geometric sequence starting with a first term of 1 and a common ratio of 10. Our task is to find the sum of the first \(n\) terms, \(S_n\), in a specific form, and then use that to derive a sum of sums. This feels like a treasure hunt through powers of 10—let’s get started!
(a) Find an Expression for \(S_n\)
A geometric sequence multiplies each term by a constant ratio. Here, the first term is 1, and the common ratio is 10, so the sequence is:
\(u_1 = 1\),
\(u_2 = 1 \cdot 10 = 10\),
\(u_3 = 10 \cdot 10 = 100\),
\(u_4 = 100 \cdot 10 = 1000\),
and so on. The \(n\)th term is \(u_n = 1 \cdot 10^{n-1} = 10^{n-1}\).
The sum of the first \(n\) terms of a geometric sequence, \(S_n\), is given by the formula:
\[
S_n = a \frac{r^n – 1}{r – 1}
\]
where \(a\) is the first term and \(r\) is the common ratio, assuming \(r \neq 1\). Here, \(a = 1\) and \(r = 10\), so:
\[
S_n = 1 \cdot \frac{10^n – 1}{10 – 1} = \frac{10^n – 1}{9}
\]
The problem asks for \(S_n\) in the form \(\frac{a^n – 1}{b}\), where \(a\) and \(b\) are positive integers. Let’s match this up:
– Numerator: \(10^n – 1\), suggesting \(a = 10\),
– Denominator: \(9\), so \(b = 9\).
Thus:
\[
S_n = \frac{10^n – 1}{9}
\]
Both 10 and 9 are in \(\mathbb{Z}^+\), . Let’s test it:
\(n = 1\): \(S_1 = \frac{10^1 – 1}{9} = \frac{10 – 1}{9} = 1\), which is the first term.
\(n = 2\): \(S_2 = \frac{10^2 – 1}{9} = \frac{100 – 1}{9} = \frac{99}{9} = 11\), and \(1 + 10 = 11\).
\(n = 3\): \(S_3 = \frac{10^3 – 1}{9} = \frac{1000 – 1}{9} = \frac{999}{9} = 111\), and \(1 + 10 + 100 = 111\).
The pattern holds—it’s the sum of \(1, 10, 100, \ldots\) up to \(n\) terms, and the formula fits perfectly.
(b) Show that \(S_1 + S_2 + S_3 + \cdots + S_n = \frac{10(10^n – 1) – 9n}{81}\)
Now we need the sum of the sums: \(S_1 + S_2 + S_3 + \cdots + S_n\). Using our expression:
– \(S_1 = \frac{10^1 – 1}{9} = \frac{9}{9} = 1\),
– \(S_2 = \frac{10^2 – 1}{9} = \frac{99}{9} = 11\),
– \(S_3 = \frac{10^3 – 1}{9} = \frac{999}{9} = 111\),
and generally, \(S_k = \frac{10^k – 1}{9}\).
We’re summing this from \(k = 1\) to \(k = n\):
\[
\sum_{k=1}^n S_k = \sum_{k=1}^n \frac{10^k – 1}{9} = \frac{1}{9} \sum_{k=1}^n (10^k – 1)
\]
Distribute the sum:
\[
= \frac{1}{9} \left( \sum_{k=1}^n 10^k – \sum_{k=1}^n 1 \right)
\]
– The first part is a geometric series: \(\sum_{k=1}^n 10^k = 10^1 + 10^2 + \cdots + 10^n\). For a geometric series \(\sum_{k=1}^n ar^{k-1}\) adjusted to start at \(k=1\) with the first term as \(10^1\), use:
\[
\sum_{k=0}^{n-1} 10 \cdot 10^k = 10 \frac{10^n – 1}{10 – 1} = 10 \frac{10^n – 1}{9}
\]
Adjusting indices, \(\sum_{k=1}^n 10^k = 10 + 10^2 + \cdots + 10^n\), which is:
\[
\frac{10(10^n – 1)}{9} \cdot \frac{10}{10} = \frac{10^{n+1} – 10}{9}
\]
– The second part: \(\sum_{k=1}^n 1 = n\), since it’s 1 added \(n\) times.
So:
\[
\sum_{k=1}^n S_k = \frac{1}{9} \left( \frac{10^{n+1} – 10}{9} – n \right)
\]
Simplify inside:
\[
= \frac{1}{9} \cdot \frac{10^{n+1} – 10 – 9n}{9} = \frac{10^{n+1} – 10 – 9n}{81}
\]
We need to match \(\frac{10(10^n – 1) – 9n}{81}\). Let’s expand the target:
\[
10(10^n – 1) – 9n = 10 \cdot 10^n – 10 – 9n = 10^{n+1} – 10 – 9n
\]
So:
\[
\frac{10(10^n – 1) – 9n}{81} = \frac{10^{n+1} – 10 – 9n}{81}
\]
………………………..Markscheme……………………………….
Ans:
(a) \(S_{n}=\frac{10^{n}-1}{9}\)
\((a=10, b=9)\)
(b) METHOD 1
\(S_{1}+S_{2}+S_{3}+…+S_{n}\)
\(=\frac{10-1}{9}+\frac{10^{2}-1}{9}+…+\frac{10^{n}-1}{9}\)
\(=\frac{10-1+10^{2}-1+10^{3}-1+…+10^{n}-1}{9}\) OR \(=\frac{9(10-1+10^{2}-1+10^{3}-1+…+10^{n}-1)}{81}\)
attempt to use geometric series formula on powers of 10, and collect -1’s together
\(10+10^{2}+10^{3}+…+10^{n}=\frac{10(10^{n}-1)}{10-1}\) and \(-1-1-1…=-n\)
\(=\frac{\frac{10(10^{n}-1)}{10-1}-n}{9}\) OR \(=\frac{9(\frac{10(10^{n}-1)}{10-1})-9n}{81}\)
\(=\frac{10(10^{n}-1)-9n}{81}\)
METHOD 2
attempt to create sum using sigma notation with Sn
\(\sum_{i=1}^{n}\frac{10^{i}-1}{9} \left ( =\frac{1}{9}\left ( \sum_{i=1}^{n}10^{i}-\sum_{i=1}^{n}1 \right ) \right )\)
\(\sum_{i=1}^{n}10^{i}=\frac{10(10^{n}-1)}{9}\)
\(\sum_{i=1}^{n}1=n\)
\(=\frac{1}{9}\left ( \frac{10\left ( 10^{n}-1 \right )}{9}-n \right )\) OR \(\frac{1}{9}\left ( \frac{10\left ( 10^{n}-1 \right )-9n}{9} \right )\)
\(=\frac{10\left ( 10^{n}-1 \right )-9n}{81}\)
METHOD 3
