Question

Consider the binomial expansion $R=(1+2x{)}^n=I+f,$ where $I$ is the integral part of $R$ and $f$ is the fractional part of $R$ for $n\in N$. If the sum of the coefficients of $R$ is $6561$ and $K=n+R-Rf,$ then which of the following is/are correct?

• A. $K=9$ when $x=\frac{1}{\sqrt{2}}$
• B. $K=7$ when $x=\frac{1}{\sqrt{2}}$
• C. $5^{\text{th}}$ term is the greatest term when $x=\frac{1}{2}$
• D. If $k^{\text{th}}$ term has the greatest coefficient, then the sum of all possible value(s) of $k$ is $13$

Answer 1

Answer: (A), (C), (D)

If $k^{\text{th}}$ term has the greatest coefficient, then the sum of all possible value(s) of $k$ is $13$

$R=(1+2x{)}^n=I+f$
Put $x=1$ to get sum of all the coefficients
$\therefore 3^n=6561=3^8\Rightarrow n=8$

For $x=\frac{1}{\sqrt{2}};R=(\sqrt{2}+1{)}^8$
Now, consider
$\underset{I+f+f^{\prime }}{\underset{}{(\sqrt{2}+1{)}^8+(\sqrt{2}-1{)}^8}}=2\left[{}^8{C}_0\right(\sqrt{2}{)}^8+....]=\text{even integer}$

Since, $I$ is integer $\Rightarrow f+f^{\prime }$ must be an integer.
But $0<f+f^{\prime }<2$
$\therefore f+f^{\prime }=1\Rightarrow f^{\prime }=1-f$

Now, $K=n+R-Rf$
$\Rightarrow K=n+R(1-f)=8+(\sqrt{2}+1{)}^n\times (\sqrt{2}-1{)}^n$
$\Rightarrow K=8+1=9$

For $R=(1+2x{)}^8,$
$T_{r+1}={}^8{C}_r(2x{)}^r$
$\Rightarrow T_{r+1}={}^8{C}_r$ when $x=\frac{1}{2}$
Now, $T_{r+1}\ge T_r$
$\Rightarrow \frac{T_{r+1}}{T_r}\ge 1$
$\Rightarrow \frac{{}^8{C}_r}{{}^8{C}_{r-1}}\ge 1$
$\Rightarrow \frac{8!}{r!(8-r)!}\times \frac{(r-1)!(9-r)!}{8!}\ge 1$
$\Rightarrow \frac{9-r}{r}\ge 1$
$\Rightarrow 9\ge 2r$
For $r=1,2,3,4$ this is true
i.e., $T_5>T_4$
But for $r=5,T_6<T_5$
$\Rightarrow T_5$ is the greatest term.


Again, $T_{k+1}={}^8{C}_k\times 2^k\times x^k;$
$T_k={}^8{C}_{k-1}\times 2^{k-1}\times x^{k-1};$
$T_{k-1}={}^8{C}_{k-2}\times 2^{k-2}\times x^{k-2}$
We have to find the term having the greatest coefficient.
$\therefore 2^{k-1}\times {}^8{C}_{k-1}\ge 2^k\times {}^8{C}_k...\left(1\right)$
And $2^{k-1}\times {}^8{C}_{k-1}\ge 2^{k-2}\times {}^8{C}_{k-2}...\left(2\right)$

From $\left(1\right),$
$\frac{8!\times 2^{k-1}}{(k-1)!(9-k)!}\ge \frac{2^k\times 8!}{k!(8-k)!}\Rightarrow \frac{1}{(9-k)}\ge \frac{2}{k}\Rightarrow k\ge 18-3k\Rightarrow k\ge 6$

Again, $2^{k-1}\times {}^8{C}_{k-1}\ge 2^{k-2}\times {}^8{C}_{k-2}$
$\Rightarrow \frac{8!\times 2^{k-1}}{(k-1)!(9-k)!}\ge \frac{2^{k-2}\times 8!}{(k-2)!(10-k)!}\Rightarrow \frac{2}{k-1}\ge \frac{1}{10-k}\Rightarrow 20-2k\ge k-1\Rightarrow k\le 7\Rightarrow 6\le k\le 7$
$\Rightarrow T_6$ and $T_7$ term have the greatest coefficient.
$\Rightarrow k=6$ or $7$
$\therefore$ Sum $=6+7=13$