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Question Number 49220 by malwaan last updated on 04/Dec/18

$${(1+\mathrm{x}-{2\mathrm{x}}^2)}^8=?$$

Commented by Abdo msup. last updated on 04/Dec/18

$$S=\sum _{k=0}^8{C}_8^k{(x-2x^2)}^k=\sum _{k=0}^8{C}_8^k{x}^k{(1-2x)}^k$$$$=\sum _{k=0}^8{C}_8^k{x}^k\left\{\sum _{p=0}^k{C}_k^p{(-2x)}^p\right\}$$$$=\sum _{k=0}^8{C}_8^k\left(\sum _{p=0}^k{(-2)}^p{C}_k^p{x}^p\right)x^k$$$$=1+C_8^1(1-2C_1^{1}x)x+C_8^2(1-2C_2^{1}x+4C_2^2{x}^2)x^2+...$$$$+C_8^8\left(\sum _{p=0}^8{(-2)}^p{C}_8^p{x}^p\right)x^8.$$

Commented by malwaan last updated on 05/Dec/18

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

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