The $n$ th term of the series
$1 + 4 + 13 + 40 + 121 + 364 + . . . . . . .$ is

\frac{1}{2} (3^{n} + 1)

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\frac{1}{2} (3^{n} - 1)

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(\frac{2^{n} + 1}{2}) 2

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3^{n} - 1

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Solution

The correct option is B $\frac{1}{2} (3^{n} - 1)$
Let $S_{n} = 1 + 4 + 13 + 40 + 121 + 364 + . . . . T_{n + 1} + T_{n}$
Rewrite $S_{n}$ again
$S_{n} = 1 + 4 + 13 + 40 + 121 + 364 + . . . . . . + T_{n - 2} + T_{n - 1} + T_{n}$
Here
$S_{n} - S_{n} = 1 + 3 + 9 + 27 + 81 + 243 + (T_{n} - T_{n - 1}) - T_{n} = 1 + 3 + 3^{2} + 3^{3} + 3^{4} + 3^{5} + (T_{n} - T_{n - 1}) - T_{n}$
$1 + 3 + 3^{2} + 3^{3} + . . . . + (T_{n} - T_{n - 1})$ is the sum of n terms of G.P.
$∴$ $n$ th term of the series
$\Rightarrow T_{n} = \frac{3^{n} - 1}{3 - 1} = \frac{3^{n} - 1}{2}$

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