If the sum of the first 10 terms of the series 4 11+4 1^4+ 4 21+4 2^4+ 4 31+4 3^4+ …. is mn, where gcd (m ,n )=1, then m+n is equal to

Question

Accepted Answer

Correct answer: 441. (441) 4.11+4.1^4+ 4.21+4.2^4+ 4.31+4.3^4+ … T_r= 4r1+4r^4= 4r4r^4+4r^2+1-4r^2 = 4r(2r^2+1)^2- (2r )^2 T_r= 4r(2r^2-2r+1)(2r^2+2r+1) T_r= (2r^2+2r+1)-(2r^2-2r+1)(2r^2-2r+1)(2r^2+2r+1) T_r=( 1r^2+ (r-1 )^2- 1r^2+ (r+1 )^2) _r=1^10 0.05emT_r=( 10^2+1^2- 11^2+2^2+ 11^2+2^2- 12^2+3^2+ … …) 19^2+10^2- 110^2+11^2 =1- 1221 = 220221 ∴m+n=220+221 =441

$If the sum of the first 10 terms of the series \frac{4 \cdot 1}{1 + 4 \cdot 1 ^{4}} + \frac{4 \cdot 2}{1 + 4 \cdot 2 ^{4}} + \frac{4 \cdot 3}{1 + 4 \cdot 3 ^{4}} + \dots . is \frac{m}{n}, where gcd (m, n) = 1, then m + n is equal to$

Solution

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