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# $a_{1}=50$ and $a_{1}+a_{2}+....+a_{n}=n^2a_{n}$ $\forall\;n \geq 1$, value of $a_{50}$ equal to

$(a)\;\large\frac{1}{51}\qquad(b)\;\large\frac{2}{51}\qquad(c)\;\large\frac{1}{101}\qquad(d)\;\large\frac{2}{101}$

Can you answer this question?

Answer : (b) $\;\large\frac{2}{51}$
Explanation : $\;a_{1}+a_{2}+......+a_{n}=n^2a_{n}$
$\;a_{1}+a_{2}+......+a_{n-1}=(n^2-1)a_{n}$
$(n-1)^2 a_{n-1}=(n^2-1)\;a_{n}$
$a_{n}=\large\frac{n-1}{n+1}\;a_{n-1}$
$a_{n}=\large\frac{(n-1)}{(n+1)}\;.\large\frac{(n-2)}{(n)}\;a_{n-2}$
$=\large\frac{(n-1)}{(n+1)}\;.\large\frac{(n-2)}{(n)}\;.\large\frac{(n-3)}{(n-1)}\;a_{n-3}$
$=\large\frac{(n-1)(n-2)(n-3)....1}{(n+1) n (n-1).....3}\;a_{1}$
$a_{n}=\large\frac{2.1}{(n+1) . n}\;a_{1}$
$a_{n}=\large\frac{2 (50)}{(n+1) . n}$
$a_{n}=\large\frac{2 (50)}{(51)(50)}$
$a_{50}=\large\frac{2}{51}\;.$
answered Jan 23, 2014 by