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Find n, if 1×2+2×3+3×4+4×5+...upto n terms1+2+3+4+...upto n terms=1003. - Mathematics and Statistics

Sum

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`.

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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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APPEARS IN

Balbharati Mathematics and Statistics 1 (Commerce) 11th Standard Maharashtra State Board
Chapter 4 Sequences and Series
Exercise 4.5 | Q 9 | Page 63
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