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Maths-

General

Easy

Question

Let $T subscript r end subscript$ and $S subscript r end subscript$ be the $r to the power of t h end exponent$ term and sum up to $r to the power of t h end exponent$ term of a series respectively. If for an odd number $n comma blank S subscript n end subscript equals n$ and $T subscript n end subscript equals fraction numerator T subscript n minus 1 end subscript over denominator n to the power of 2 end exponent end fraction$ , then $T subscript m end subscript$ ( $m$ being even) is

$\frac{2}{1 + m^{2}}$
$\frac{2 m^{2}}{1 + m^{2}}$
$\frac{{(m + 1)}^{2}}{2 + {(m + 1)}^{2}}$
$\frac{2 {(m + 1)}^{2}}{1 + {(m + 1)}^{2}}$

The correct answer is: $fraction numerator 2 open parentheses m plus 1 close parentheses to the power of 2 end exponent over denominator 1 plus open parentheses m plus 1 close parentheses to the power of 2 end exponent end fraction$

$S subscript n end subscript minus S subscript n minus 2 end subscript equals 2$
$rightwards double arrow T subscript n end subscript plus T subscript n minus 1 end subscript equals 2$
Also, $T subscript n end subscript plus T subscript n minus 1 end subscript equals open parentheses fraction numerator 1 over denominator n to the power of 2 end exponent end fraction plus 1 close parentheses T subscript n minus 1 end subscript equals 2$
$rightwards double arrow T subscript n minus 1 end subscript equals fraction numerator 2 over denominator 1 plus fraction numerator 1 over denominator n to the power of 2 end exponent end fraction end fraction equals fraction numerator 2 n to the power of 2 end exponent over denominator 1 plus n to the power of 2 end exponent end fraction$
So, $T subscript m end subscript equals fraction numerator 2 open parentheses m plus 1 close parentheses to the power of 2 end exponent over denominator 1 plus open parentheses m plus 1 close parentheses to the power of 2 end exponent end fraction$

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