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JEE Class main Answered

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Asked by swayamagarwal2114 | 09 Aug, 2022, 01:44: PM
Expert Answer
Let space straight a subscript straight k equals 3 to the power of straight k over straight k
rightwards double arrow straight S equals sum from straight m equals 1 to infinity of space sum from straight n equals 1 to infinity of space fraction numerator straight m squared straight n over denominator left parenthesis 3 to the power of straight m right parenthesis left parenthesis straight n 3 to the power of straight m space plus space straight m 3 to the power of straight n right parenthesis end fraction equals sum from straight m equals 1 to infinity of space sum from straight m equals 1 to infinity of space fraction numerator 1 over denominator straight a subscript straight m left parenthesis straight a subscript straight m plus straight a subscript straight n right parenthesis end fraction space space space... space left parenthesis straight i right parenthesis
Interchanging space straight m space and space straight n
rightwards double arrow straight S equals sum from straight m equals 1 to infinity of space sum from straight n equals 1 to infinity of space fraction numerator 1 over denominator straight a subscript straight n left parenthesis straight a subscript straight n plus straight a subscript straight m right parenthesis end fraction space space space... space left parenthesis ii right parenthesis
From space left parenthesis straight i right parenthesis space & space left parenthesis ii right parenthesis comma space we space get
2 straight S equals sum from straight m equals 1 to infinity of space sum from straight n equals 1 to infinity of space fraction numerator 1 over denominator straight a subscript straight m space straight a subscript straight n end fraction equals open parentheses sum from straight k equals 1 to infinity of 1 over straight a subscript straight k close parentheses squared equals space open parentheses sum from straight k equals 1 to infinity of space straight k over 3 to the power of straight k close parentheses squared equals open parentheses sum from straight k equals 1 to infinity of space straight k 3 to the power of negative straight k end exponent close parentheses squared
Since comma space sum from straight k equals 1 to infinity of space straight k space straight x to the power of straight k space equals space straight x over left parenthesis 1 minus straight x right parenthesis squared comma space straight x less than 1
rightwards double arrow straight S equals 1 half open parentheses 3 over left parenthesis 1 minus 3 right parenthesis squared close parentheses equals 1 half cross times 9 over 16 equals 9 over 32
Answered by Renu Varma | 29 Aug, 2022, 10:18: AM
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