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Maths-

General

Easy

Question

The sum to 50 terms of the series $fraction numerator 3 over denominator 1 to the power of 2 end exponent end fraction plus fraction numerator 5 over denominator 1 to the power of 2 end exponent plus 2 to the power of 2 end exponent end fraction plus fraction numerator 7 over denominator 1 to the power of 2 end exponent plus 2 to the power of 2 end exponent plus 3 to the power of 2 end exponent end fraction plus horizontal ellipsis$ is

$\frac{100}{17}$
$\frac{150}{17}$
$\frac{200}{51}$
$\frac{50}{17}$

The correct answer is: $fraction numerator 100 over denominator 17 end fraction$

Let $T subscript r end subscript$ be the $r to the power of t h end exponent$ term of the given series. Then,
$T subscript r end subscript equals fraction numerator 2 r plus 1 over denominator 1 to the power of 2 end exponent plus 2 to the power of 2 end exponent plus horizontal ellipsis plus r to the power of 2 end exponent end fraction$
$equals fraction numerator 6 left parenthesis 2 r plus 1 right parenthesis over denominator open parentheses r close parentheses open parentheses r plus 1 close parentheses left parenthesis 2 r plus 1 right parenthesis end fraction$
$equals 6 open parentheses fraction numerator 1 over denominator r end fraction minus fraction numerator 1 over denominator r plus 1 end fraction close parentheses$
So, sum is given by
$not stretchy sum from r equals 1 to 50 of T subscript r end subscript equals 6 not stretchy sum from r equals 1 to 50 of open parentheses fraction numerator 1 over denominator r end fraction minus fraction numerator 1 over denominator r plus 1 end fraction close parentheses$
$equals 6 open square brackets open parentheses 1 minus fraction numerator 1 over denominator 2 end fraction close parentheses plus open parentheses fraction numerator 1 over denominator 2 end fraction minus fraction numerator 1 over denominator 3 end fraction close parentheses plus open parentheses fraction numerator 1 over denominator 3 end fraction minus fraction numerator 1 over denominator 4 end fraction close parentheses plus horizontal ellipsis plus open parentheses fraction numerator 1 over denominator 50 end fraction minus fraction numerator 1 over denominator 51 end fraction close parentheses close square brackets$
$equals 6 open square brackets 1 minus fraction numerator 1 over denominator 51 end fraction close square brackets$
$equals fraction numerator 100 over denominator 17 end fraction$

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