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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`.
Solution
`(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
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