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Question:

The sum 3 × 1³1² + 5 × (1³ + 2³)1² + 2³ + 7 × (1³ + 2³ + 3³)1² + 2² + 3² + ...

620

660

680

600

Solution:

Correct option is A.660Tn=(3+(n−1;)×2)(13+23+...+n(12+22+...+n2)=32n(n+1)=n(n+1)(n+2)−(n−1;)n(n+1)2⇒Sn=n(n+1)(n+2)2⇒S10=660