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


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

$F i n d t h e s u m o f t h e c o e f f i c i e n t s$ $o f a l l t h e i n t e g r a l p o w e r o f x i n t h e$ $e x p a n s i o n o f {(1 + 2 \sqrt{x})}^{40} .$
	Answered by MJS last updated on 22/Mar/18
	$(1+2(√x))^2 =4x+1+... (1+2(√x))^4 =16x^2 +24x+1+... (1+2(√x))^6 =64x^3 +240x^2 +60x+1+... {4;1}={1×2^2 ;1×2^0 } {16;24;1}={1×2^4 ;6×2^2 ;1×2^0 } {64;240;60;1}= ={1×2^6 ;15×2^4 ;15×2^2 ;1×2^0 } we see: { ((n),(0) )×2^n ; ((n),(2) )×2^(n−2) ; ((n),(4) )×2^(n−4) ;...} the sum of this list is S=Σ_(k=0) ^(n/2) ((n),((2k)) )×2^(n−2k) ; with 2∣n n=40 ⇒ S=6 078 832 729 528 464 401$
	${(1 + 2 \sqrt{x})}^{2} = 4 x + 1 + . . .$ ${(1 + 2 \sqrt{x})}^{4} = 16 x^{2} + 24 x + 1 + . . .$ ${(1 + 2 \sqrt{x})}^{6} = 64 x^{3} + 240 x^{2} + 60 x + 1 + . . .$ ${4; 1} = {1 \times 2^{2}; 1 \times 2^{0}}$ ${16; 24; 1} = {1 \times 2^{4}; 6 \times 2^{2}; 1 \times 2^{0}}$ ${64; 240; 60; 1} =$ $= {1 \times 2^{6}; 15 \times 2^{4}; 15 \times 2^{2}; 1 \times 2^{0}}$ $we see :$ ${(\begin{matrix} n \\ 0 \end{matrix}) \times 2^{n}; (\begin{matrix} n \\ 2 \end{matrix}) \times 2^{n - 2}; (\begin{matrix} n \\ 4 \end{matrix}) \times 2^{n - 4}; . . .}$ $the sum of this list is$ $S = \underset{k = 0}{\sum^{\frac{n}{2}}} (\begin{matrix} n \\ 2 k \end{matrix}) \times 2^{n - 2 k}; with 2 ∣ n$ $n = 40$ $\Rightarrow S = 6 078 832 729 528 464 401$
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