A) \[\frac{m(m+1)}{2}\]
B) \[\frac{m(m+1)(2m+1)}{6}\]
C) \[\frac{n(n+1)(2n+1)}{6}\]
D) \[\frac{n(n+1)}{2}\]
Correct Answer: C
Solution :
It is nothing but\[{{S}_{\infty }}=\frac{a}{1-r}=\frac{\sqrt{2}+1}{1-(\sqrt{2}-1)}\].
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