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

Concept: Observe the pattern: \[ 1=2^1-1 \] \[ 3=2^2-1 \] \[ 7=2^3-1 \] \[ 15=2^4-1 \] Hence the \(r\)-th term is \[\begin{aligned} t_r=2^r-1 \end{aligned}\]

Step 1: Write the sum to \(n\) terms. \[\begin{aligned} S_n &= \sum_{r=1}^{n}(2^r-1) \\ &= \sum_{r=1}^{n}2^r - \sum_{r=1}^{n}1 \end{aligned}\]

Step 2: Evaluate each sum. \[\begin{aligned} \sum_{r=1}^{n}2^r = 2^{n+1}-2 \end{aligned}\] and \[\begin{aligned} \sum_{r=1}^{n}1=n \end{aligned}\] Therefore, \[\begin{aligned} S_n &= (2^{n+1}-2)-n \end{aligned}\] \[\begin{aligned} &= 2^{n+1}-2-n \end{aligned}\]

Step 3: Write the final answer. \[\begin{aligned} \boxed{ 2^{n+1}-2-n } \end{aligned}\] Hence, option \(\mathbf{(A)}\) is correct.