InterviewSolution
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InterviewSolution
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lt 1. +22 +32 +(A) π2/12upto oo-6, then 1a +32 +52 +(B) π32416J=(C) π78(D) π2/4PART - II: SINGLE AND DOUBLE VALUE INTEGER TYPE |
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Answer» 1/1² + 1/3² + 1/5² + ...= (1/1² + 1/2² + 1/3² + 1/4² + 1/5² + 1/6² + ...) - (1/2² + 1/4² + 1/6² + ...)= (1/1² + 1/2² + 1/3² + ...) - (1/2²)(1 + 1/2² + 1/3² + ...)= (1 - 1/2²)(1 + 1/2² + 1/3² + ...)= (3/4)(π²/6)= π²/8. |
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