Question

The sum to \(n\) terms of the series \[ 1+3+7+15+\cdots \] is

• A. \(2^{n+1}-2-n\)
• B. \(2^{n+1}-2\)
• C. \(2^{n+1}-2-n^2\)
• D. \(2^{n+1}-n\)

Hint:

Whenever a series looks like \[ 1,\;3,\;7,\;15,\ldots \] rewrite each term as \[ 2^r-1 \] and then use the geometric progression sum formula.

Answer 1

The Correct Option is A

Step 1: Spot the pattern in the terms.
The terms are $1, 3, 7, 15, \ldots$ Each one is just one less than a power of $2$: $1 = 2-1$, $3 = 4-1$, $7 = 8-1$, $15 = 16-1$.

Step 2: Write the general term.
So the $r$-th term is $t_r = 2^r - 1$.

Step 3: Split the sum into two simpler sums.
$S_n = \sum_{r=1}^{n}(2^r - 1) = \sum_{r=1}^{n} 2^r - \sum_{r=1}^{n} 1$.

Step 4: Add up the powers of 2.
The first part is a geometric series: $2 + 4 + 8 + \cdots + 2^n = 2^{n+1} - 2$.

Step 5: Add up the ones.
The second part is simply $n$, since we add $1$ a total of $n$ times.

Step 6: Combine the two results.
So $S_n = (2^{n+1} - 2) - n = 2^{n+1} - 2 - n$. \[ \boxed{2^{n+1} - 2 - n} \]