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1 + 4 + 13 + 40 + 121 + ... - Mathematics

1 + 4 + 13 + 40 + 121 + ...

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Solution

Let \[T_n\] be the nth term and \[S_n\] be the sum to n terms of the given series.
Thus, we have:

\[S_n = 1 + 4 + 13 + 40 + 121 + . . . + T_{n - 1} + T_n\]   ...(1)

Equation (1) can be rewritten as:

\[S_n = 1 + 4 + 13 + 40 + 121 + . . . + T_{n - 1} + T_n\] ...(2)

On subtracting (2) from (1), we get:

\[S_n = 1 + 4 + 13 + 40 + 121 + . . . + T_{n - 1} + T_n \]

\[ S_n = 1 + 4 + 13 + 40 + 121 + . . . + T_{n - 1} + T_n \]

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\[ 0 = 1 + \left[ 3 + 9 + 27 + 81 + . . . + \left( T_n - T_{n - 1} \right) \right] - T_n\]

The sequence of difference between successive terms is 3, 9, 27, 81,...
We observe that it is a GP with common ratio 3 and first term 3.
Thus, we have:

\[1 + \left[ \frac{3\left( 3^{n - 1} - 1 \right)}{3 - 1} \right] - T_n = 0\]

\[ \Rightarrow 1 + \left[ \frac{\left( 3^n - 3 \right)}{2} \right] - T_n = 0\]

\[ \Rightarrow \left( \frac{3^n}{2} - \frac{1}{2} \right) - T_n = 0\]

\[ \Rightarrow \left( \frac{3^n}{2} - \frac{1}{2} \right) = T_n\]

\[\because S_n = \sum^n_{k = 1} T_k \]

\[ \therefore S_n = \sum^n_{k = 1} \left( \frac{3^k}{2} - \frac{1}{2} \right)\]

\[ \Rightarrow S_n = \frac{1}{2} \sum^n_{k = 1} 3^k - \frac{1}{2} \sum^n_{k = 1} 1\]

\[ \Rightarrow S_n = \frac{1}{2}\left( 3 + 3^2 + 3^3 + 3^4 + 3^5 + . . . + 3^n \right) - \frac{n}{2}\]

\[ \Rightarrow S_n = \frac{1}{2}\left[ \frac{3\left( 3^n - 1 \right)}{2} \right] - \frac{n}{2}\]

\[ \Rightarrow S_n = \left( \frac{3^{n + 1} - 3}{4} \right) - \frac{n}{2}\]

\[ \Rightarrow S_n = \left( \frac{3^{n + 1} - 3 - 2n}{4} \right)\]

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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 21 Some special series
Exercise 21.2 | Q 6 | Page 18
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