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The sum of first 9 terms of the series 1³/1 +1³+2³/1+3...... Is
[hint: find the nth term and then find its summation]Tn for the series is : Tn= {(n+1)^2 }/4Then use the summation of this Tn with n from 1 to 9. ∑ Tn = 96.Hope this helps. Comment if you didn’t understand the above solution.Anuj
HereTr= ((n+1))^2/4that implies sum of nine digits will be = (1/4)(1^2+2^2+3^2--------10^2-1^2)=96
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