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EXERCISE 2.6 · Q84

Q.Find ∑r=1n(r+1)(2r−1)\displaystyle\sum_{r=1}^{n}(r+1)(2r-1).

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✓ Free question

(r+1)(2r−1)=2r2+r−1(r+1)(2r-1)=2r^2+r-1. ∑r=1n(2r2+r−1)=2⋅n(n+1)(2n+1)6+n(n+1)2−n\sum_{r=1}^n(2r^2+r-1)=2\cdot\dfrac{n(n+1)(2n+1)}6+\dfrac{n(n+1)}2-n. Combining over a common denominator of 6: =2n(n+1)(2n+1)+3n(n+1)−6n6=n[2(n+1)(2n+1)+3(n+1)−6]6=n(4n2+9n−1)6=\dfrac{2n(n+1)(2n+1)+3n(n+1)-6n}6=\dfrac{n[2(n+1)(2n+1)+3(n+1)-6]}6=\dfrac{n(4n^2+9n-1)}6 (check n=1n=1: term =2=2, formula gives 1(12)/6=21(12)/6=2 ✓).

✓Final answer

Sn=n(4n2+9n−1)6S_n=\dfrac{n(4n^2+9n-1)}6.

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