The sum of 10 terms of the series 31^22^2+52^23^2+73^24^2+ is:

Question

The sum of 10 terms of the series $\dfrac{3}{1^2	imes2^2}+\dfrac{5}{2^2	imes3^2}+\dfrac{7}{3^2	imes4^2}+\cdots$ is:

Accepted Answer

$$S = {{{2^2} - {1^2}} \over {{1^2} \times {2^2}}} + {{{3^2} - {2^2}} \over {{2^2} \times {3^2}}} + {{{4^2} - {3^2}} \over {{3^2} \times {4^2}}} + ...$$

$$ = \left[ {{1 \over {{1^2}}} - {1 \over {{2^2}}}} \right] + \left[ {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right] + \left[ {{1 \over {{3^2}}} - {1 \over {{4^2}}}} \right] + .... + \left[ {{1 \over {{{10}^2}}} - {1 \over {{{11}^2}}}} \right]$$

$$ = 1 - {1 \over {121}}$$

$$ = {{120} \over {121}}$$

Solutions · 1