Find-the-sum-of-the-coefficients-of-all-the-integral-power-of-x-in-the-expansion-of-1-2-x-40-

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 \sqrt{x})}^{2} = 4 x + 1 + \dots

{(1 + 2 \sqrt{x})}^{4} = 16 x^{2} + 24 x + 1 + \dots

{(1 + 2 \sqrt{x})}^{6} = 64 x^{3} + 240 x^{2} + 60 x + 1 + \dots

{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}; \dots}

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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