The n-th term of the given series is:
Tn=n1(12+22+…+n2)
Using the formula for the sum of squares of the first n natural numbers:
Tn=n1(6n(n+1)(2n+1))=6(n+1)(2n+1)=62n2+3n+1
The sum of the first 10 terms is S10=n=1∑10Tn
S10=61(2n=1∑10n2+3n=1∑10n+n=1∑101)
Using the standard summation formulas:
n=1∑10n2=610×11×21=385
n=1∑10n=210×11=55
n=1∑101=10
Substituting these values into the sum expression:
S10=61(2×385+3×55+10)
S10=61(770+165+10)
S10=6945=2315
Answer: 2315