Sequence And Series Question 391
Question: The value of $ \sum\limits_{n=1}^{10}{\sum\limits_{m=1}^{10}{(m^{2}+n^{2})}} $ equals
Options:
A) 4235
B) 5050
C) 7700
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ \sum\limits_{n=1}^{10}{\sum\limits_{m=1}^{10}{(m^{2}+n^{2})}} $ $ =\sum\limits_{n=1}^{10}{[(1^{2}+n^{2})+(2^{2}+n^{2})+….+(10^{2}+n^{2})]} $ $ =10[{{(1)}^{2}}+{{(2)}^{2}}+….+{{(10)}^{2}}]+10[{{(1)}^{2}} $ $ +{{(2)}^{2}}+….+{{(10)}^{2}}] $ $ =\frac{20.10.11.21}{6}=7700 $
BETA
