Question

Find n, if `(1 xx 2 + 2 xx 3 + 3 xx 4 + 4 xx 5 + ... + "upto n terms")/(1 + 2 + 3 + 4 + ... + "upto n terms")= 100/3`.

Answer 1

`(1 xx 2 + 2 xx 3 + 3 xx 4 + 4 xx 5 + ... + "upto n terms")/(1 + 2 + 3 + 4 + ... + "upto n terms")= 100/3`

∴ \[\frac{\displaystyle\sum_{r=1}^{n} r(r + 1)}{\displaystyle\sum_{r=1}^{n} r} = \frac{100}{3}\]

∴ \[\frac{\displaystyle\sum_{r=1}^{n} r^2 + \displaystyle\sum_{r=1}^{n} r}{\displaystyle\sum_{r=1}^{n} r} = \frac{100}{3}\]

∴ `(("n"("n" + 1)(2"n" + 1))/6 + ("n"("n" + 1))/2)/(("n"("n" + 1))/2) = 100/3`

∴  `(("n"("n" + 1))/6[(2"n" + 1) + 3])/(("n"("n" + 1))/2) = 100/3`

∴ `(2("n" + 2))/3 = 100/3`

∴ n + 2 = 50

∴ n = 48