Consider two G.Ps. 2,2^{2}, 2^{3}, ldots and | vedclass

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Question

$\left(\left(2^{1} 2^{2} \cdots 2^{60}\right)\left(4^{1} \cdot 4^{2} \ldots \ldots 4^{ n }\right)\right)^{\frac{1}{60+ n }}=2^{\frac{225}{8}}$

$\le

Vedclass · Accepted Answer

$1330$

Vedclass · Answer

$\left(\left(2^{1} 2^{2} \cdots 2^{60}ight)\left(4^{1} \cdot 4^{2} \ldots \ldots 4^{ n }ight)ight)^{\frac{1}{60+ n }}=2^{\frac{225}{8}}$

$\left(2^{30 	imes 61} 4^{\frac{ n ( n +1)}{2}}ight) \frac{1}{60+ n }=2^{\frac{225}{8}}$

$2^{1830+ n ^{2}+ n }=2^{\frac{(225)(60+ n )}{8}}$

$=8 n ^{2}-217 n +1140=0$

$n =20, \frac{57}{8}$

$\sum_{ k =1}^{ n } nk - k ^{2}=\frac{ n ^{2}( n +1)}{2}-\frac{ n ( n +1)(2 n +1)}{6}$

$=1330$

Consider two G.Ps. $2,2^{2}, 2^{3}, \ldots$ and $4,4^{2}, 4^{3}, \ldots$ of $60$ and $n$ terms respectively. If the geometric mean of all the $60+n$ terms is $(2)^{\frac{225}{8}}$, then $\sum_{ k =1}^{ n } k (n- k )$ is equal to.

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