Consider two G.Ps. 2,2^2,2^3,… and 4,4^2,4^3,… of 60 and n terms respectively. If the geometric mean of all the 60+n terms is (2)^ 225 8, then Σ…

Given two G.Ps. $2,{2}^{2},{2}^{3},\ldots$2,22,23,… and $4,{4}^{2},{4}^{3},\ldots$4,42,43,… of $60$60 and $n$n terms respectively,

Also given the geometric mean of all the $60+n$60+n terms is ${(2)}^{\frac{225}{8}}$(2)8225​,

So, ${(({2}^{1}{2}^{2}\ldots .{2}^{60})({4}^{1}\cdot {4}^{2}\ldots \ldots .{4}^{n}))}^{\frac{1}{60+n}}={2}^{\frac{225}{8}}$((2122….260)(41⋅42…….4n))60+n1​=28225​

$$\Rightarrow {({2}^{30\times 61}{4}^{\frac{n(n+1)}{2}})}^{\frac{1}{60+n}}={2}^{\frac{225}{8}}$$⇒(230×6142n(n+1)​)60+n1​=28225​

$$\Rightarrow {2}^{1830+{n}^{2}+n}={2}^{\frac{(225)(60+n)}{8}}$$⇒21830+n2+n=28(225)(60+n)​

On comparing both side we get,

$$\Rightarrow 1830+{n}^{2}+n=\frac{225(60+n)}{8}$$⇒1830+n2+n=8225(60+n)​

$$\Rightarrow 8{n}^{2}-217n+1140=0$$⇒8n2−217n+1140=0

$$\Rightarrow n=20,\frac{57}{8}$$⇒n=20,857​

Now $\displaystyle \sum _{k=1}^{n}k(n-k)=$k=1∑n​k(n−k)=$\displaystyle \sum _{k=1}^{20}nk-{k}^{2}=\frac{{n}^{2}(n+1)}{2}-\frac{n(n+1)(2n+1)}{6}$k=1∑20​nk−k2=2n2(n+1)​−6n(n+1)(2n+1)​

$$=\frac{{20}^{2}\times 21}{2}-\frac{20\times 21\times 41}{6}=1330$$=2202×21​−620×21×41​=1330