If 115+125+135+145+........... upto infinite terms = k, then value of ( 115+135+155+175+........... upto infinite terms) will be

S=115+135+155+175+.............=(115+125+135+145+........)−(125+145+165+.......)=(k)−132(115+125+135+...........)=k−k32=31k32

If 115+125+135+145+........... upto infinite terms = k, then value of ( 115+135+155+175+........... upto infinite terms) will be

S=115+135+155+175+.............=(115+125+135+145+........)−(125+145+165+.......)=(k)−132(115+125+135+...........)=k−k32=31k32

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If $\frac{1}{1^{5}} + \frac{1}{2^{5}} + \frac{1}{3^{5}} + \frac{1}{4^{5}} + ...........$ upto infinite terms = k, then value of ( $\frac{1}{1^{5}} + \frac{1}{3^{5}} + \frac{1}{5^{5}} + \frac{1}{7^{5}} + ...........$ upto infinite terms) will be

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$\frac{31 k}{32}$

$\frac{k}{32}$

$\frac{k}{16}$

$\frac{15 k}{16}$

answer is B.

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$S = \frac{1}{1^{5}} + \frac{1}{3^{5}} + \frac{1}{5^{5}} + \frac{1}{7^{5}} + .............$

$= (\frac{1}{1^{5}} + \frac{1}{2^{5}} + \frac{1}{3^{5}} + \frac{1}{4^{5}} + ........) - (\frac{1}{2^{5}} + \frac{1}{4^{5}} + \frac{1}{6^{5}} + .......)$

$= (k) - \frac{1}{32} (\frac{1}{1^{5}} + \frac{1}{2^{5}} + \frac{1}{3^{5}} + ...........) = k - \frac{k}{32} = \frac{31 k}{32}$

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If 115+125+135+145+........... upto infinite terms = k, then value of ( 115+135+155+175+........... upto infinite terms) will be

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If $\frac{1}{1^{5}} + \frac{1}{2^{5}} + \frac{1}{3^{5}} + \frac{1}{4^{5}} + ...........$ upto infinite terms = k, then value of ( $\frac{1}{1^{5}} + \frac{1}{3^{5}} + \frac{1}{5^{5}} + \frac{1}{7^{5}} + ...........$ upto infinite terms) will be