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Question Number 177597 by cortano1 last updated on 07/Oct/22

Answered by mr W last updated on 07/Oct/22

$$\mathit{m}\mathit{e}\mathit{t}\mathit{h}\mathit{o}\mathit{d}1:\mathit{w}\mathit{i}\mathit{t}\mathit{h}\mathit{o}\mathit{u}\mathit{t}\mathit{u}\mathit{s}\mathit{i}\mathit{n}\mathit{g}\mathit{f}\mathit{o}\mathit{r}\mathit{m}\mathit{u}\mathit{l}\mathit{a}$$$$1\times 1\times C_3^6\times 2^3$$$$+[1\times (-1)+3\times 1]\times C_2^6\times 2^4$$$$+3\times (-1)\times 6\times 2^5$$$$=64$$

Answered by mr W last updated on 07/Oct/22

$$\mathit{m}\mathit{e}\mathit{t}\mathit{h}\mathit{o}\mathit{d}2:\mathit{u}\mathit{s}\mathit{i}\mathit{n}\mathit{g}\mathit{f}\mathit{o}\mathit{r}\mathit{m}\mathit{u}\mathit{l}\mathit{a}$$$$(1+3x)(1-x){(2+x)}^6$$$$=(1+2x-3x^2)\underset{k=0}{\sum ^6}{C}_k^6{2}^{6-k}{x}^k$$$$coef.of{x}^{3}term:$$$$1\times C_3^6\times 2^{6-3}+2\times C_2^6\times 2^{6-2}-3\times C_1^6\times 2^{6-1}$$$$=160+480-576$$$$=64$$

Commented by Tawa11 last updated on 09/Oct/22

$$\mathrm{Great}\mathrm{sir}.$$

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