The sum 1+ 1^3+2^31+2+ 1^3+2^3+3^31+2+3+ … …+ 1^3+2^3+3^3+ …+15^31+2+3+ …+15- 12 (1+2+3+ …+15 is equal to :

Question

Accepted Answer

Correct answer: A. (a) JEE Main 2019 Let, S=1+ 1^3+2^31+2+ 1^3+2^3+3^31+2+3+ …15 terms T_n= 1^3+2^3+ …n^31+2+ …n= ( n (n+1 )2 )^2 n (n+1 )2= n (n+1 )2 Now, S= 12( 0.05emn^2+ 0.05emn)= 12( 15 (16 ) (31 )6+ 15 (16 )2) =680 ∵ required sum is 680- 12 15 (16 )2=680-60=620

$The sum 1 + \frac{1 ^{3} + 2 ^{3}}{1 + 2} + \frac{1 ^{3} + 2 ^{3} + 3 ^{3}}{1 + 2 + 3} + \dots \dots + \frac{1 ^{3} + 2 ^{3} + 3 ^{3} + \dots + 15 ^{3}}{1 + 2 + 3 + \dots + 15} - \frac{1}{2} (1 + 2 + 3 + \dots + 15 is equal to :$

Solution

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