If S-1-12-1-22-1-32-1-42- to - then find 1-12-1-32-1-52-1-72- to - in terms of S-

S=((1/1^2)+(1/3^2)+(1/5^2)+.............∞ #10;)+((1/2^2)+(1/4^2)+......∞ ) S=x+(1/4)[(1/1^2)+(1/2^2)+(1/3^2)+.......∞ #10;] S=x+(1/4)× s ∴ ...

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Byju's Answer

Standard XII

Mathematics

Summation by Sigma Method

If S=1/12+1...

Question

If $S = \frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + . . . . . . . . . . . . .$ to $\infty$ then find $\frac{1}{1^{2}} + \frac{1}{3^{2}} + \frac{1}{5^{2}} + \frac{1}{7^{2}} + . . . . . . . . . . . . .$ to $\infty$ in terms of S.

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Solution

$S = ((\frac{1}{1^{2}}) + (\frac{1}{3^{2}}) + (\frac{1}{5^{2}}) + . . . . . . . . . . . . . \infty) + ((\frac{1}{2^{2}}) + (\frac{1}{4^{2}}) + . . . . . . \infty) S = x + (\frac{1}{4}) [(\frac{1}{1^{2}}) + (\frac{1}{2^{2}}) + (\frac{1}{3^{2}}) + . . . . . . . \infty] S = x + (\frac{1}{4}) \times s ∴ x = S - (\frac{S}{4}) = (\frac{35}{4})$

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Similar questions

Q. If

x = \frac{1}{1^{2}} + \frac{1}{3^{2}} + \frac{1}{5^{2}} + . . . .,

y = \frac{1}{1^{2}} + \frac{3}{2^{2}} + \frac{1}{3^{2}} + \frac{3}{4^{2}} + . .

and

z = \frac{1}{1^{2}} - \frac{1}{2^{2}} + \frac{1}{3^{2}} - \frac{1}{4^{2}} + . . . .,

then

Q. If

\frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + . . . . \infty = \frac{π^{2}}{6}

then,

\frac{1}{1^{2}} + \frac{1}{3^{2}} + \frac{1}{5^{2}} + . . . . \infty

is equal to

If $\frac{1}{1^{2}}$ + $\frac{1}{2^{2}}$ + $\frac{1}{3^{2}}$ + $\frac{1}{4^{2}}$ + ........ = $\frac{π^{2}}{6}$ then $\frac{1}{1^{2}}$ + $\frac{1}{3^{2}}$ + $\frac{1}{5^{2}}$ + ....... =

Q. If

\frac{1}{2^{2}} + \frac{1}{4^{2}} + \frac{1}{6^{2}} + . . . . . . \infty = \frac{π^{2}}{24}

, then

\frac{1}{1^{2}} + \frac{1}{3^{2}} + \frac{1}{5^{2}} + . . . . . . \infty

is equal to

\frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + . . . . u p t o \infty = \frac{π^{2}}{6}

, then find

\frac{1}{1^{2}} + \frac{1}{3^{2}} + \frac{1}{5^{2}} + . . . . u p t o \infty

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If S=112+122+132+142+............. to∞ then find 112+132+152+172+............. to∞ in terms of S.

S=((112)+(132)+(152)+.............∞)+((122)+(142)+......∞)S=x+(14)[(112)+(122)+(132)+.......∞]S=x+(14)×s∴x=S−(S4)=(354)

If $S = \frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + . . . . . . . . . . . . .$ to $\infty$ then find $\frac{1}{1^{2}} + \frac{1}{3^{2}} + \frac{1}{5^{2}} + \frac{1}{7^{2}} + . . . . . . . . . . . . .$ to $\infty$ in terms of S.