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Asked by swayamagarwal2114 | 09 Aug, 2022, 13:44: PM
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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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JEE main - Maths
the sum of infinite series 1/2(1/3+1/4)-1/4(1/3^2+1/4^2)+1/6(1/3^3+1/4^3)+
Asked by pritamlovevanshi5 | 15 Sep, 2024, 08:24: AM
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JEE main - Maths
s=1+1/3+1/9+1/27 value of s is
Asked by vishalsarap0022 | 19 May, 2024, 10:30: AM
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JEE main - Maths
23,47,1p156 relation
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JEE main - Maths
solv
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Asked by vishwajeetsinghrajput2018 | 17 Mar, 2024, 00:27: AM
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JEE main - Maths
solve
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Asked by mailtome.aj86 | 26 Feb, 2024, 20:29: PM
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JEE main - Maths
the value 21/4. 41/8. 81/6..... is
Asked by sabeenashaik575 | 09 Jan, 2024, 09:53: AM
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JEE main - Maths
solution
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Asked by 20sanjaykoley20 | 19 Dec, 2023, 00:57: AM
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JEE main - Maths
Let S be the sum P the product and R the sum of reciprocal
Asked by kunmunkumar08032008 | 08 Oct, 2023, 13:15: PM
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JEE main - Maths
Let Tr be the rth term of an AP for r=1,2,3,.... If for some positive integers m, n. We have Tm=1/n and Tn=1/m, then Tmn equals to
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Asked by nitinsolanki14102001 | 16 Sep, 2023, 15:01: PM
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JEE main - Maths
if an= -2/(4n^2)-(16n)+(15),      [(4n^2)-(16n)+(15) dis entire thing is the denominator)] then: opt a) a1+a2+....+a10= 20/51 opt b) a1+a2+.....a10=30/51 opt c) a1+a2+......a25=50/141 opt d) a1+a2+.....a25=60/141 Its more than one ans type q, pls tell me a detailed explanation on how to do it as i din understand 
Asked by joanmaria916 | 08 Jul, 2023, 16:37: PM
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