If S_n is the sum of n terms of a geometric p | vedclass

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Question

(C) The sum of $n$ terms of a geometric progression is given by $S_n = \frac{a(1 - r^n)}{1 - r}$.We need to find the sum $\sum_{k=1}^{n} S_{2k-1}$.Sub

Vedclass · Accepted Answer

$\frac{a}{1 - r} \left[ n - r \cdot \frac{1 - r^{2n}}{1 - r^2} \right]$

Vedclass · Answer

(C) The sum of $n$ terms of a geometric progression is given by $S_n = \frac{a(1 - r^n)}{1 - r}$.We need to find the sum $\sum_{k=1}^{n} S_{2k-1}$.Substituting the formula for $S_{2k-1}$:$\sum_{k=1}^{n} S_{2k-1} = \sum_{k=1}^{n} \frac{a(1 - r^{2k-1})}{1 - r} = \frac{a}{1 - r} \sum_{k=1}^{n} (1 - r^{2k-1})$.This can be split into two sums: $\frac{a}{1 - r} \left[ \sum_{k=1}^{n} 1 - \sum_{k=1}^{n} r^{2k-1} \right]$.The first part is $\sum_{k=1}^{n} 1 = n$.The second part is a geometric series with first term $r$ and common ratio $r^2$: $r + r^3 + r^5 + \dots + r^{2n-1} = \frac{r(1 - (r^2)^n)}{1 - r^2} = \frac{r(1 - r^{2n})}{1 - r^2}$.Thus,the total sum is $\frac{a}{1 - r} \left[ n - \frac{r(1 - r^{2n})}{1 - r^2} \right]$.

If $S_n$ is the sum of $n$ terms of a geometric progression with first term $a$ and common ratio $r$,what is the sum of $S_1 + S_3 + S_5 + \dots + S_{2n-1}$?

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