Sequence And Series Question 118
Question: The sum to $ n $ terms of the series $ 2^{2}+4^{2}+6^{2}+……….. $ is
[MP PET 1994]
Options:
A) $ \frac{n(n+1)(2n+1)}{3} $
B) $ \frac{2n(n+1)(2n+1)}{3} $
C) $ \frac{n(n+1)(2n+1)}{6} $
D) $ \frac{n(n+1)(2n+1)}{9} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ 2^{2}+4^{2}+6^{2}+……..+{{(2n)}^{2}} $ $ =2^{2}(1^{2}+2^{2}+3^{2}+…….+n^{2}) $ $ =\frac{4n(n+1)(2n+1)}{6}=\frac{2n(n+1)(2n+1)}{3} $ .
BETA
