If 1 +(1/22) + (1/32) +(1/42) + ......∞ = (π2/6)then the value of 1 -(1/22) +(1/32) - (1/42) + (1/52) - (1/62) + ......∞ equals to

Question

If 1 +(1/22) + (1/32) +(1/42) + ......∞ = (π2/6)then the value of 1 -(1/22) +(1/32) - (1/42) + (1/52) - (1/62) + ......∞ equals to (A) (π2/24) (B) (π2/48) (C) (π2/12) (D) (5π2/12)

Tardigrade Admin · Accepted Answer

Given, 1 +(1/22) + (1/32) + (1/42) + ......∞ = (π2/6) =A (say) ...(i) ∴ 1 -(1/22) + (1/32) - (1/42) + .......∞ = (1 +(1/22) + (1/32) +(1/42) + .....∞) -2 ((1/22) + (1/42) +(1/62) + ......∞) = (1 +(1/22) + (1/32) + ....∞) - (2/22) (1 +(1/22) + (1/32) + .....∞) =A - (1/2)A = (A/2) = (π2/12) (Using (i))

Q. If $1 + \frac{1}{2 ^{2}} + \frac{1}{3 ^{2}} + \frac{1}{4 ^{2}} + ......\infty = \frac{π ^{2}}{6}$ then the value of $1 - \frac{1}{2 ^{2}} + \frac{1}{3 ^{2}} - \frac{1}{4 ^{2}} + \frac{1}{5 ^{2}} - \frac{1}{6 ^{2}} + ......\infty$ equals to

$\frac{π ^{2}}{24}$

$\frac{π ^{2}}{48}$

$\frac{π ^{2}}{12}$

$\frac{5 π ^{2}}{12}$

Q. If 1+221​+321​+421​+......∞=6π2​then the value of 1−221​+321​−421​+521​−621​+......∞ equals to

24π2​

48π2​

12π2​

125π2​

Given, 1+221​+321​+421​+......∞=6π2​=A(say)...(i) ∴1−221​+321​−421​+.......∞ =(1+221​+321​+421​+.....∞)−2(221​+421​+621​+......∞) =(1+221​+321​+....∞)−222​(1+221​+321​+.....∞) =A−21​A=2A​=12π2​ (Using (i))

Q. If $1 + \frac{1}{2 ^{2}} + \frac{1}{3 ^{2}} + \frac{1}{4 ^{2}} + ......\infty = \frac{π ^{2}}{6}$ then the value of $1 - \frac{1}{2 ^{2}} + \frac{1}{3 ^{2}} - \frac{1}{4 ^{2}} + \frac{1}{5 ^{2}} - \frac{1}{6 ^{2}} + ......\infty$ equals to

$\frac{π ^{2}}{24}$

$\frac{π ^{2}}{48}$

$\frac{π ^{2}}{12}$

$\frac{5 π ^{2}}{12}$