The sum of \(n\) terms of the series \(1+\frac{3}{2}+2+\frac{5}{2}+3+\frac{7}{2}+......\)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\frac{n(n+3)}{4}\)
Each term is \(\frac{1}{2}\) more than the one before (1, 1.5, 2, 2.5, 3, …), so the series is an arithmetic progression, and the sum of \(n\) terms is \(S_n=\frac{n}{2}\left[2a+(n-1)d\right]\).
First term \(a=1\), common difference \(d=\frac{1}{2}\)
\(\displaystyle S_n=\frac{n}{2}\left[2+\frac{n-1}{2}\right]\)
\(\displaystyle S_n=\frac{n}{2}\times\frac{n+3}{2}=\frac{n(n+3)}{4}\)
Shortcut: Test the options with small values of \(n\).
\(n=1\):
The sum is 1
Options (3) and (4) give 1.5 and 2, so they are out
\(n=2\):
The sum is \(1+1.5=2.5\)
Option (1) gives 3, while Option (2) gives \(\frac{2\times5}{4}=2.5\)