Question

Let $a=(4^{1/401}-1)$ and let $b_n{=}_1^C{+}^n{C}_2.a{+}^n{C}_3.a^2+............{+}^n{C}_n.A^{n-1}$. Find the value of $(b_{2006}-b_{2005})$.

Answer 1

Given $a=(4^{1/401}-1)$

$b_n=n_{C_1}+n_{C_2}a+n_{C_3}{a}^2+......+n_{C_n}{a}^{n-1}$

$\Rightarrow {ab}_n=a^n{c}_1+n_{C_2}{a}^2+n_{C_3}{a}^3+......+n_{C_n}{a}^n$

$\Rightarrow 1+{ab}_n=n_{C_0}+n_{C_1}a+n_{C_2}{a}^2+......+n_{C_n}{a}^n={(1+a)}^n$

$\therefore$ $1+{ab}_n={(1+a)}^n={(1+4^{1/401}-1)}^n=4^{\frac{n}{401}}$

$\Rightarrow {ab}_n=4^{\frac{n}{401}}-1$

$\Rightarrow b_n=\frac{4^{\frac{n}{401}}-1}{4^{\frac{1}{401}}-1}$

$\therefore$ $b_{2006}=\frac{4^{2006/401}-1}{4^{1/401}-1}{b}_{2005}=\frac{4^{2005/401}-1}{4^{1/401}-1}=\frac{4^5-1}{4^{1/401}-1}$

$\therefore$ $b_{2006}-b_{2005}=\frac{4^{2006/401}-4^5}{4^{1/401}-1}=\frac{4^{(2005+1)/401}-4^5}{4^{1/401}-1}$

$=\frac{4^{\frac{2005}{401}}.4^{\frac{1}{401}}-4^5}{4^{1/401}-1}=\frac{4^5.4^{1/401}-4^5}{4^{1/401}-1}$

$=4^5.\left[\frac{4^{1/401}-1}{4^{1/401}-1}\right]=4^5=1024$.