Sequence And Series Question 132
Question: Sum of the squares of first $ n $ natural numbers exceeds their sum by 330, then $ n= $
[Karnataka CET 1998]
Options:
A) 8
B) 10
C) 15
D) 20
Show Answer
Answer:
Correct Answer: B
Solution:
$ \Sigma n^{2}=330+\Sigma n $
$ \Rightarrow $ $ \frac{n,(n+1),(2n+1)}{6}=330+\frac{n,(n+1)}{2} $
Þ $ \frac{n,(n+1)}{2}[ \frac{2n+1}{3}-1 ]=330 $
Þ $ \frac{n,(n+1)}{2}\ .\ \frac{2(n-1)}{3}=330 $
Þ $ n,(n+1)(n-1)=990 $
$ \Rightarrow $ $ n=10 $ .
BETA
