The sum of 50 terms of 3-12-5-12-22-7-12-22-32- - is

The correct option is B t_r = 2r+1/1^2 + 2^2 +.... + r^2 = 6/r(r + 1) ∑_r = 1^50 t_r = 6 ∑_r = 1^501/r(r+1) = 6 ∑_r = 1^50 ( 1/r 1/r + 1 ) = 6 ( 1 1/51 ) ...

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Byju's Answer

Standard XII

Mathematics

General Term of Binomial Expansion

The sum of 50...

Question

The sum of 50 terms of $\frac{3}{1^{2}} + \frac{5}{1^{2} + 2^{2}} + \frac{7}{1^{2} + 2^{2} + 3^{2}} + . . . .$ is

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Solution

The correct option is B
$t_{r} = \frac{2 r + 1}{1^{2} + 2^{2} + . . . . + r^{2}} = \frac{6}{r (r + 1)} \sum_{r = 1}^{50} t_{r} = 6 \sum_{r = 1}^{50} \frac{1}{r (r + 1)} = 6 \sum_{r = 1}^{50} (\frac{1}{r} - \frac{1}{r + 1}) = 6 (1 - \frac{1}{51}) = \frac{100}{17}$

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Q. The sum

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The sum of 50 terms of 312+512+22+712+22+32+.... is

The correct option is B tr=2r+112+22+....+r2=6r(r+1)∑50r=1tr=6∑50r=11r(r+1)=6∑50r=1(1r−1r+1)=6(1−151)=10017

The sum of 50 terms of $\frac{3}{1^{2}} + \frac{5}{1^{2} + 2^{2}} + \frac{7}{1^{2} + 2^{2} + 3^{2}} + . . . .$ is

The correct option is B
$t_{r} = \frac{2 r + 1}{1^{2} + 2^{2} + . . . . + r^{2}} = \frac{6}{r (r + 1)} \sum_{r = 1}^{50} t_{r} = 6 \sum_{r = 1}^{50} \frac{1}{r (r + 1)} = 6 \sum_{r = 1}^{50} (\frac{1}{r} - \frac{1}{r + 1}) = 6 (1 - \frac{1}{51}) = \frac{100}{17}$