Question

Find the sum to $n$ terms of the series, $1+\left(1+\frac{1}{2}+\frac{1}{2^2}\right)+\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}\right)+\dots$

Answer 1

2n - \frac{4}{3} + \frac{1}{3 \cdot 4^{n-1}}

Answer 2

The $k$-th term of the series, $T_k$, is identified as a geometric progression with first term 1, common ratio 1/2, and $2k-1$ terms. Its sum is found using the GP sum formula to be $T_k = 2 - \frac{1}{2^{2k-2}}$. The sum of the series to $n$ terms, $S_n = \sum_{k=1}^{n} T_k$, is then calculated. This sum simplifies to $2n - \sum_{k=1}^{n} \frac{1}{4^{k-1}}$. The second part is another geometric progression with first term 1, common ratio 1/4, and $n$ terms. Its sum is $\frac{4}{3} - \frac{1}{3 \cdot 4^{n-1}}$. Substituting this back yields the final expression for $S_n$.