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Question Number 32239 by rahul 19 last updated on 22/Mar/18

Find the sum of the coefficients of all the integral power of x in the expansion of (1+2(√x))^(40) .

$$\boldsymbol{{F}}{ind}\:{the}\:{sum}\:{of}\:{the}\:{coefficients} \\ $$$${of}\:{all}\:{the}\:{integral}\:{power}\:{of}\:{x}\:{in}\:{the} \\ $$$${expansion}\:{of}\:\left(\mathrm{1}+\mathrm{2}\sqrt{{x}}\right)^{\mathrm{40}} . \\ $$

Answered by MJS last updated on 22/Mar/18

(1+2(√x))^2 =4x+1+... (1+2(√x))^4 =16x^2 +24x+1+... (1+2(√x))^6 =64x^3 +240x^2 +60x+1+... {4;1}={1×2^2 ;1×2^0 } {16;24;1}={1×2^4 ;6×2^2 ;1×2^0 } {64;240;60;1}= ={1×2^6 ;15×2^4 ;15×2^2 ;1×2^0 } we see: { ((n),(0) )×2^n ; ((n),(2) )×2^(n−2) ; ((n),(4) )×2^(n−4) ;...} the sum of this list is S=Σ_(k=0) ^(n/2) ((n),((2k)) )×2^(n−2k) ; with 2∣n n=40 ⇒ S=6 078 832 729 528 464 401

$$\left(\mathrm{1}+\mathrm{2}\sqrt{{x}}\right)^{\mathrm{2}} =\mathrm{4}{x}+\mathrm{1}+... \\ $$$$\left(\mathrm{1}+\mathrm{2}\sqrt{{x}}\right)^{\mathrm{4}} =\mathrm{16}{x}^{\mathrm{2}} +\mathrm{24}{x}+\mathrm{1}+... \\ $$$$\left(\mathrm{1}+\mathrm{2}\sqrt{{x}}\right)^{\mathrm{6}} =\mathrm{64}{x}^{\mathrm{3}} +\mathrm{240}{x}^{\mathrm{2}} +\mathrm{60}{x}+\mathrm{1}+... \\ $$$$\left\{\mathrm{4};\mathrm{1}\right\}=\left\{\mathrm{1}×\mathrm{2}^{\mathrm{2}} ;\mathrm{1}×\mathrm{2}^{\mathrm{0}} \right\} \\ $$$$\left\{\mathrm{16};\mathrm{24};\mathrm{1}\right\}=\left\{\mathrm{1}×\mathrm{2}^{\mathrm{4}} ;\mathrm{6}×\mathrm{2}^{\mathrm{2}} ;\mathrm{1}×\mathrm{2}^{\mathrm{0}} \right\} \\ $$$$\left\{\mathrm{64};\mathrm{240};\mathrm{60};\mathrm{1}\right\}= \\ $$$$=\left\{\mathrm{1}×\mathrm{2}^{\mathrm{6}} ;\mathrm{15}×\mathrm{2}^{\mathrm{4}} ;\mathrm{15}×\mathrm{2}^{\mathrm{2}} ;\mathrm{1}×\mathrm{2}^{\mathrm{0}} \right\} \\ $$$$ \\ $$$$\mathrm{we}\:\mathrm{see}: \\ $$$$\left\{\begin{pmatrix}{{n}}\\{\mathrm{0}}\end{pmatrix}×\mathrm{2}^{{n}} ;\begin{pmatrix}{{n}}\\{\mathrm{2}}\end{pmatrix}×\mathrm{2}^{{n}−\mathrm{2}} ;\begin{pmatrix}{{n}}\\{\mathrm{4}}\end{pmatrix}×\mathrm{2}^{{n}−\mathrm{4}} ;...\right\} \\ $$$$\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{this}\:\mathrm{list}\:\mathrm{is} \\ $$$${S}=\underset{{k}=\mathrm{0}} {\overset{\frac{{n}}{\mathrm{2}}} {\sum}}\begin{pmatrix}{{n}}\\{\mathrm{2}{k}}\end{pmatrix}×\mathrm{2}^{{n}−\mathrm{2}{k}} ;\:\mathrm{with}\:\mathrm{2}\mid{n} \\ $$$${n}=\mathrm{40} \\ $$$$\Rightarrow\:{S}=\mathrm{6}\:\mathrm{078}\:\mathrm{832}\:\mathrm{729}\:\mathrm{528}\:\mathrm{464}\:\mathrm{401} \\ $$

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