Consider two G.P.s 2,2^2,2^3, (60 terms) and 4,4^2,4^3, (n terms).

Question

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

Accepted Answer

The GM equation gives $\dfrac{\frac{60\times61}{2}+2\cdot\frac{n(n+1)}{2}}{60+n} = \dfrac{225}{8}$, leading to $8n^2-217n+1140=0$, whose valid root is $n=20$.

$$\sum_{k=1}^{20}k(20-k) = 20\sum k - \sum k^2 = 20\left(\frac{20\times21}{2}\right)-\frac{20\times21\times41}{6} = 4200-2870 = 1330.$$

Solutions · 1