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Question Number 40254 by scientist last updated on 17/Jul/18

$$sumofthisseries:$$$$1.3.5+3.5.7+5.7.9+...uptonterms$$

Commented by maxmathsup by imad last updated on 18/Jul/18

$$\sum _{k=0}^n(2k+1)(2k+3)(2k+5)=1.3.5+3.5.7+5.7.9+....+(2n+1)(2n+3)(2n+5)$$$$=S_n$$$$S_n=\sum _{k=0}^n(4k^2+6k+2k+3)(2k+5)$$$$=\sum _{k=0}^n(4k^2+8k+3)(2k+5)$$$$=\sum _{k=0}^n(8k^3+20k^2+16k^2+40k+6k+15)$$$$=\sum _{k=0}^n(8k^3+32k^2+46k+15)$$$$=8\sum _{k=0}^n{k}^3+32\sum _{k=0}^n{k}^2+46\sum _{k=0}^{n}k+15\sum _{k=0}^n\left(1\right)$$$$=8\frac{n^2{(n+1)}^2}{4}+32\frac{n(n+1)(2n+1)}{6}+46\frac{n(n+1)}{2}+15n$$$$=2n^2{(n+1)}^2+\frac{16}{3}n(n+1)(2n+1)+23n(n+1)+15n.$$

Commented by scientist last updated on 18/Jul/18

$$but{can}^{\prime }tweuseTn=n(n+2)(n+4)$$

Commented by maxmathsup by imad last updated on 18/Jul/18

$$nolooksir\sum _{k=1}^n{T}_k=\sum _{k=1}^{n}k(k+2)(k+4)=1.3.4+2.4.6+...$$$$andthissumistotallydifferntfromthesumgiveninthequestion...$$

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