The sum of all the coefficients in the binomial expansion of \((1+2 x)^n\) is 6561. Let \(R=(I+2 x)^n=I+F\),…
The sum of all the coefficients in the binomial expansion of \((1+2 x)^n\) is 6561. Let \(R=(I+2 x)^n=I+F\), where \(I \in N\) and \(0 < F < \mathrm{l}\).
If \(x=\frac{1}{\sqrt{2}}\), then \(1-\frac{F}{1+(\sqrt{2}-1)^4}=\)
\((3 \sqrt{2}-4)\)
\(4(3 \sqrt{2}+4)\)
\((\sqrt{2}-1)^4\)
1
Solution
It is given that sum of all the coefficients in the binomial expansion of \((1+2 x)^n\) is \(6561=(1+2)^n\) on putting \(x=1\).
\(\begin{array}{ll}
\Rightarrow & 3^n=6561 \\
\Rightarrow & n=8
\end{array}\)
Now, at \(x=\frac{1}{\sqrt{2}}\), then \(R=(1+2 x)^n=I+F\)
\(\begin{aligned}
& \Rightarrow R=(\sqrt{2}+1)^8=I+F, \text { where } I \in N \text { and } 0 < F < 1 \\
& \therefore(\sqrt{2}-1)^8=F^{\prime}, \text { where } 0 < F^{\prime} < 1 \\
& \therefore(\sqrt{2}+1)^8+(\sqrt{2}-1)^8=I+\left(F+F^{\prime}\right) \\
& \Rightarrow 2\left[(\sqrt{2})^8+{ }^8 C_2(\sqrt{2})^6+{ }^8 C_4(\sqrt{2})^4+{ }^8 C_6(\sqrt{2})^2+{ }^8 C_8\right] \\
& \quad=I+\left(F+F^{\prime}\right) \\
& \Rightarrow \text { Even integer }=I+\left(F+F^{\prime}\right) \\
& \Rightarrow F+F^{\prime} \in \text { Integer } \\
& \because 0 < F < 1 \text { and } 0 < F^{\prime} < 1 \Rightarrow 0 < F+F^{\prime} < 2 \\
& \text { So, } F+F^{\prime}=1 \\
& \Rightarrow F=1-F^{\prime}=1-(\sqrt{2}-1)^8
\end{aligned}\)
\(\begin{aligned}
& \text { So, } 1-\frac{F}{1+(\sqrt{2}-1)^4}=1-\frac{1-(\sqrt{2}-1)^8}{1+(\sqrt{2}-1)^4} \\
& =1-\frac{\left[1+(\sqrt{2}-1)^4\right]\left[1-(\sqrt{2}-1)^4\right]}{1+(\sqrt{2}-1)^4} \\
& =1-\left[1-(\sqrt{2}-1)^4\right]=(\sqrt{2}-1)^4
\end{aligned}\)
Hence, option (3) is correct.