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

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(C) The product of the terms of the first $G$.$P$. is $P_1 = 2^{1+2+\dots+60} = 2^{\frac{60 	imes 61}{2}} = 2^{1830}$.The product of the terms of the

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$1330$

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(C) The product of the terms of the first $G$.$P$. is $P_1 = 2^{1+2+\dots+60} = 2^{\frac{60 	imes 61}{2}} = 2^{1830}$.The product of the terms of the second $G$.$P$. is $P_2 = 4^{1+2+\dots+n} = 4^{\frac{n(n+1)}{2}} = 2^{n(n+1)}$.The geometric mean of all $60+n$ terms is given by $(P_1 	imes P_2)^{\frac{1}{60+n}} = 2^{\frac{225}{8}}$.$(2^{1830} 	imes 2^{n^2+n})^{\frac{1}{60+n}} = 2^{\frac{225}{8}}$.$2^{\frac{n^2+n+1830}{60+n}} = 2^{\frac{225}{8}}$.Equating the exponents: $\frac{n^2+n+1830}{60+n} = \frac{225}{8}$.$8n^2 + 8n + 14640 = 13500 + 225n$.$8n^2 - 217n + 1140 = 0$.Solving for $n$: $(n-20)(8n-57) = 0$. Since $n$ must be an integer,$n=20$.We need to calculate $\sum_{k=1}^{n} k(n-k) = n \sum_{k=1}^{n} k - \sum_{k=1}^{n} k^2$.$= n \frac{n(n+1)}{2} - \frac{n(n+1)(2n+1)}{6} = \frac{n(n+1)}{6} [3n - (2n+1)] = \frac{n(n+1)(n-1)}{6} = \frac{n(n^2-1)}{6}$.For $n=20$: $\frac{20(400-1)}{6} = \frac{20 	imes 399}{6} = 10 	imes 133 = 1330$.

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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