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  4. If (1/2 × 3 × 4)+(1/3 × 4 × 5)+(1/4 × 5 × 6)+ ldots+ (1/100 × 101 × 102)=( k /101), then 34 k is equal to

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Q. If $\frac{1}{2 \times 3 \times 4}+\frac{1}{3 \times 4 \times 5}+\frac{1}{4 \times 5 \times 6}+\ldots+$ $\frac{1}{100 \times 101 \times 102}=\frac{ k }{101}$, then $34 k$ is equal to

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Solution:

$ \frac{1}{2.3 .4}+\frac{1}{3.4 .5}+\ldots . .+\frac{1}{100.101 .102}=\frac{ k }{101} $
$ \frac{4-2}{2.3 .4}+\frac{5-3}{3.4 .5}+\ldots . .+\frac{102-100}{100.101 .102}=\frac{2 k }{101}$
$ \frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+\ldots . .+\frac{1}{100.101}-\frac{1}{101.102}=\frac{2 k }{101}$
$ \frac{1}{2.3}-\frac{1}{101.102}=\frac{2 k }{101}$
$ \therefore 2 k =\frac{101}{6}-\frac{1}{102} $
$ \therefore 34 k =286$