Sequence And Series Question 117
Question: The sum of first $ n $ terms of the given series $ 1^{2}+{{2.2}^{2}}+3^{2}+{{2.4}^{2}}+5^{2}+{{2.6}^{2}}+………… $ is $ \frac{n{{(n+1)}^{2}}}{2} $ , when $ n $ is even. When $ n $ is odd, the sum will be
[IIT 1988]
Options:
A) $ \frac{n{{(n+1)}^{2}}}{2} $
B) $ \frac{1}{2}n^{2}(n+1) $
C) $ n{{(n+1)}^{2}} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
When $ n $ is odd, the last term. $ i.e., $ the $ n^{th} $ term will be $ n^{2} $ in this case $ n-1 $ is even and so the sum of the first $ n-1 $ terms of the series is obtained by replacing $ n $ by $ n-1 $ in the given formula and so is $ \frac{1}{2}(n-1)n^{2} $ . Hence the sum of the n terms = (the sum of $ n-1 $ terms) + the $ n^{th} $ term $ =\frac{1}{2}(n-1)n^{2}+n^{2}=\frac{1}{2}(n+1)n^{2} $ . Trick: Check for $ n=1,\ 3 $ . Here $ S_1=1,\ S_3=18 $ which gives (b).
BETA
