Find the Sum of the Series 1 + 3 + 7 + 15 + ⋯ up to n Terms

A standard sequences and series problem, not tied to a specific exam paper.

Problem

Find the sum of the series $1+3+7+15+\cdots$ up to $n$ terms.

Key insight. The series itself is neither an AP nor a GP, but the differences between consecutive terms — $2,4,8,\ldots$ — are a clean geometric progression. Writing the sum shifted by one position and subtracting it from itself is what turns those differences into the sum that needs to be evaluated.

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Approach

Write the sum $S$ of the first $n$ terms, then write it again shifted one position to the right, and subtract. Every term except the first and last cancels in pairs, leaving a new sum made entirely of the differences between consecutive original terms — which form a geometric progression. This reveals the $n$-th term $T_n$ explicitly; summing $T_n$ over $n=1$ to $n$ (itself a GP minus a constant series) gives the final answer.

Solution

Step 1 — Set up the method of differences

Let $S = T_1+T_2+\cdots+T_{n-1}+T_n = 1+3+7+15+\cdots+T_n$.

Write it again, shifted by one term:

$$S = \phantom{1+{}}1+3+7+\cdots+T_{n-1}+T_n$$

Subtracting the shifted version from the original (aligning terms) leaves:

$$T_n = 1 + (3-1)+(7-3)+(15-7)+\cdots+(T_n-T_{n-1})$$

$$T_n = 1 + 2+4+8+\cdots \text{ up to } (n-1) \text{ terms}$$

Step 2 — Recognise the geometric series

The terms $2,4,8,\ldots$ (there are $n-1$ of them) form a GP with first term $2$ and common ratio $2$:

$$2+4+8+\cdots \text{($n-1$ terms)} = \frac{2(2^{n-1}-1)}{2-1} = 2^n-2$$

Step 3 — Write the general term

$$T_n = 1+(2^n-2) = 2^n-1$$

Check: $T_1=2^1-1=1$ ✓, $T_2=2^2-1=3$ ✓, $T_3=2^3-1=7$ ✓, $T_4=2^4-1=15$ ✓.

Step 4 — Sum the general term from 1 to n

$$S_n = \sum_{k=1}^n T_k = \sum_{k=1}^n (2^k-1) = \sum_{k=1}^n 2^k – \sum_{k=1}^n 1$$

Step 5 — Evaluate each piece

The first sum is a GP: $2+4+8+\cdots+2^n = 2(2^n-1) = 2^{n+1}-2$.

The second sum is simply $n$ (adding $1$, $n$ times).

Step 6 — Combine

$$S_n = (2^{n+1}-2) – n = 2^{n+1}-n-2$$

Answer

Common mistakes

• Stopping at the general term $T_n=2^n-1$ without summing it. Finding $T_n$ is the middle step, not the final answer — the question asks for the sum of the series, which needs one more summation over $k=1$ to $n$.
• Losing track of the “$n-1$ terms” count in the differences. Since $S$ shifted by one position aligns $n-1$ pairs of terms (not $n$), the geometric series in Step 2 has $n-1$ terms, going up to $2^{n-1}$ — miscounting this shifts every subsequent power of 2 by one.
• Writing $2^n-2$ instead of $2^{n+1}-2$ for the GP sum in Step 5. The sum $2+4+\cdots+2^n$ has $n$ terms ending at $2^n$, and its total is $2^{n+1}-2$ — it’s easy to mistakenly write the last term’s own value ($2^n$) as if it were the sum.

Practise next

• Find the sum of $2+5+11+23+\cdots$ up to $n$ terms.