The sum of \(n\) terms of the series \(1+1.5+2+2.5+3+\dots\) is?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{n(n+3)}{4}\)
Each term is 0.5 more than the one before, so the series is an arithmetic progression with first term \(a=1\) and common difference \(d=0.5\). The sum of its first \(n\) terms is \(S_n=\frac{n}{2}\left[2a+(n-1)d\right]\).
\(\displaystyle S_n=\frac{n}{2}\left[2+(n-1)\times\frac{1}{2}\right]\)
\(\displaystyle S_n=\frac{n}{2}\times\frac{4+n-1}{2}\)
\(\displaystyle S_n=\frac{n(n+3)}{4}\)
Shortcut: Test small values of \(n\) in the options.
\(n=1\): the sum is 1; Options (3) and (4) give 1.5 and 2, so they go
\(n=2\): the sum is \(1+1.5=2.5\); Option (1) gives \(\frac{2\times5}{4}=2.5\), Option (2) gives \(\frac{2\times3}{2}=3\)