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EXERCISE 2.6 · Q86

Q.Find ∑r=1n1+2+3+⋯+rr\displaystyle\sum_{r=1}^{n}\dfrac{1+2+3+\cdots+r}{r}.

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✓ Free question

1+2+⋯+r=r(r+1)21+2+\cdots+r=\dfrac{r(r+1)}2, so the term is r(r+1)/2r=r+12\dfrac{r(r+1)/2}r=\dfrac{r+1}2. Sn=∑r=1nr+12=12[n(n+1)2+n]=n(n+3)4S_n=\sum_{r=1}^n\dfrac{r+1}2=\dfrac12\left[\dfrac{n(n+1)}2+n\right]=\dfrac{n(n+3)}4 (check n=1n=1: term =1=1, formula gives 1(4)/4=11(4)/4=1 ✓).

✓Final answer

Sn=n(n+3)4S_n=\dfrac{n(n+3)}4.

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