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Question Number 37270 by abdo.msup.com last updated on 11/Jun/18

$$find{\int }_0^1{\left(\frac{1-x^{n+1}}{1-x}\right)}^{2}dx.$$

Answered by abdo.msup.com last updated on 27/Jul/18

$$letI={\int }_0^1{\left(\frac{1-x^{n+1}}{1-x}\right)}^{2}dx$$$$I={\int }_0^1{(1+x+x^2+...+x^n)}^{2}dx$$$$={\int }_0^1{\left(\sum _{i=0}^n{x}^i\right)}^{2}dx$$$$={\int }_0^1(\sum _{i=0}^n{x}^{2i}+2\sum _{1⩽i<j⩽n}{x}^{i+j})dx$$$$=\sum _{i=0}^n\frac{1}{2i+1}+2\sum _{1⩽i<j⩽n}\frac{1}{i+j}$$

Commented by abdo.msup.com last updated on 27/Jul/18

$$I=\sum _{i=0}^n\frac{1}{2i+1}+2\sum _{1⩽i<j⩽n}\frac{1}{i+j+1}$$

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