The sum of all the coefficients in the binomi | vedclass

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Question

(C) The sum of all coefficients in $(1+2x)^n$ is obtained by setting $x=1$,which gives $(1+2)^n = 3^n = 6561$. Since $3^8 = 6561$,we have $n=8$. Given

Vedclass · Accepted Answer

$(\sqrt{2}-1)^4$

Vedclass · Answer

(C) The sum of all coefficients in $(1+2x)^n$ is obtained by setting $x=1$,which gives $(1+2)^n = 3^n = 6561$. Since $3^8 = 6561$,we have $n=8$. Given $R = (1+2x)^n = (1+\sqrt{2})^8 = I+F$,where $I \in N$ and $0 Let $F' = (\sqrt{2}-1)^8$. Since $0 Consider $R + F' = (\sqrt{2}+1)^8 + (\sqrt{2}-1)^8$. Expanding using the binomial theorem,the odd terms cancel out,resulting in an even integer. Thus,$I+F+F' = 	ext{Even Integer}$,which implies $F+F' = 1$ because $0 Therefore,$F = 1 - F' = 1 - (\sqrt{2}-1)^8$. Now,calculate $1 - \frac{F}{1+(\sqrt{2}-1)^4} = 1 - \frac{1-(\sqrt{2}-1)^8}{1+(\sqrt{2}-1)^4}$. Using the difference of squares $a^2-b^2 = (a-b)(a+b)$,we have $1-(\sqrt{2}-1)^8 = [1-(\sqrt{2}-1)^4][1+(\sqrt{2}-1)^4]$. Substituting this,we get $1 - [1-(\sqrt{2}-1)^4] = (\sqrt{2}-1)^4$. Thus,the correct option is $C$.

The sum of all the coefficients in the binomial expansion of $(1+2x)^n$ is $6561$. Let $R=(1+2x)^n=I+F$,where $I \in N$ and $0 < F < 1$. If $x=\frac{1}{\sqrt{2}}$,then $1-\frac{F}{1+(\sqrt{2}-1)^4}=$

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