If - n - 55 then value of - n2 is equal to

The value of ${lim}_{n \to \infty} [\frac{n}{1 + n^{2}} + \frac{n}{4 + n^{2}} + \frac{n}{9 + n^{2}} + \dots + \frac{1}{2 n}]$ is equal to [Bihar CEE 1994]

Q. If

S_{1} = \sum n, S_{2} = \sum n^{2}

and

S_{3} = \sum n^{3}

, then the value of

{lim}_{n \to \infty} \frac{S_{1} (1 + S_{3} / 8)}{S_{2}^{2}}

is equal to

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If ∑n=55 then value of ∑n2 is equal to

The correct option is D 385∑n=55⇒n(n+1)2=55⇒n2+n−110=0⇒n=10,n≠−11∴∑n2=n(n+1)(2n+1)6=10×11×216=385Hence choice (a) is correct answer.

If $\sum n = 55$ then value of $\sum n^{2}$ is equal to

The correct option is D $385$
$\sum n = 55$
$\Rightarrow \frac{n (n + 1)}{2} = 55$
$\Rightarrow n^{2} + n - 110 = 0 \Rightarrow n = 10, n \neq - 11$
$∴ \sum n^{2} = \frac{n (n + 1) (2 n + 1)}{6} = \frac{10 \times 11 \times 21}{6} = 385$
Hence choice (a) is correct answer.