131+(13+23)2+(13+23+33)3+....uptonterms=
n(n+1)(n+2)(5n+3)48
n(n+1)(n+2)(n+3)24
n(n+1)(n+2)(7n+1)48
n(n+1)(n+2)(3n+5)48
Explanation for the correct option:
Step 1. Find the nthterm of the given series:
Let Sn=131+(13+23)2+(13+23+33)3+.... up to n terms
nth term of given series is Tn=13+23+33+…n3n
=n(n+1)22n=n(n+1)24 ∵∑n3=(n(n+1)2)2
=n(n2+2n+1)4=(n3+2n2+n)4
Step 2. Find the sum of the series:
Sn=∑Tn
=∑(n3+2n2+n)4
Using formulae,
=n(n+1)224+2n(n+1)(2n+1)64+n(n+1)24 [∑n3=(nn+12)2;∑n2=n(n+1)(2n+1)6;∑n=n(n+1)2]
=n(n+1)2n(n+1)4+4(2n+1)3+12
=n(n+1)43n2+11n+1012
=n(n+1)43n2+5n+6n+1012
=n(n+1)4n(3n+5)+2(3n+5)12
=n(n+1)(n+2)(3n+5)48
Hence, Option ‘D’ is Correct.
If the sum of squares of the intercept on the axes cut off by the tangent on the curve x13+y13=a13,a>0 at (a8,a8) is 2, then value of a is