Question

Help me

Answer 1

The sum of coefficients of all integral powers in the expansion of
\[\huge (1+2\sqrt{x})^{40}\]

Answer 2

@mathmate @zepdrix @mathmath333 @Miracrown @jim_thompson5910

Answer 3

here u have to look the pattern
\(\large\tt \begin{align} \color{black}{(1+2\sqrt{x})^1=1+2=3=3^1\\~\\
(1+2\sqrt{x})^2=1+2+2=9=3^2\\~\\
(1+2\sqrt{x})^3=1+6+8+12=27=3^3\\~\\
(1+2\sqrt{x})^4=1+8+16+24+32=81=3^4\\~\\
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\Large (1+2\sqrt{x})^{40}=.............=3^{40}
}\end{align}\)

Answer 4

Also notice the pattern:
\(\large \sum_{i=0,2}^2 (^2_i)2^i=(9+1)/2=(3^2+1)/2\)
\(\large \sum_{i=0,2}^3 (^3_i)2^i=(27+1)/2=(3^3+1)/2\)
\(\large \sum_{i=0,2,4}^4 (^4_i)2^i=(81+1)/2=(3^4+1)/2\)
...
\(\large \sum_{i=0,2,4,6,...}^{40} (^{40}_i)2^i=?\)