Question

The coefficient of three consecutive terms in the expansion of $(1+x{)}^n$ are in the ratio 1 : 7 : 42. Find r and n. Also, determine the square root of n + 1-r.

Or

Find the coefficient of the term independent of x in the expansion of ${(\frac{x+1}{x^{\frac{2}{3}}-x^{\frac{1}{3}}+1}-\frac{x-1}{x-x^{\frac{1}{2}}})}^{10}$.

Answer 1

Let the three consecutive terms be rth, (r + 1)th and (r+2)th terms. Then, their coefficients in the expansion of $(1+x{)}^n$ are ${}^n{C}_{r-1}$, ${}^n{C}_r$ and ${}^n{C}_{r+1}$ respectively.

It is given that, ${}^n{C}_{r-1}{:}^n{C}_r:{}^n{C}_{r+1}=1:7:42$

Now, $\frac{{}^n{C}_{r-1}}{{}^n{C}_r}=\frac{1}{7}\Rightarrow \frac{r}{n-r+1}=\frac{1}{7}[\because \frac{{}^n{C}_r}{{}^n{C}_r}=\frac{n-r+1}{r}]$

$\Rightarrow n-8r+1=0$ .....(i)

and $\frac{{}^n{C}_r}{{}^n{C}_{r+1}}=\frac{7}{42}\Rightarrow \frac{r+1}{n-r}=\frac{1}{6}$ $\Rightarrow n-7r-6=0$ .....(ii)

On solving Eqs. (i) and (ii), we get r = 7 and n = 55 Now, $\sqrt{n+1-r}=\sqrt{55+1-7}=\sqrt{49}=7$

Or

We have, $\frac{x+1}{x^{\frac{2}{3}}-x^{\frac{1}{3}}+1}-\frac{x-1}{x-x^{\frac{1}{2}}}=\frac{(x^{\frac{1}{3}}{)}^3+(1{)}^3}{x^{\frac{2}{3}}-x^{\frac{1}{3}}+1}-\frac{(x^{\frac{1}{2}}{)}^2-(1{)}^2}{x^{\frac{1}{2}}(x^{\frac{1}{2}}-1)}$

$=\frac{(x^{\frac{1}{3}}+1)(x^{\frac{2}{3}}-x^{\frac{1}{3}}+1)}{x^{\frac{2}{3}}-x^{\frac{1}{3}}+1}-\frac{(x^{\frac{1}{2}}+1)(x^{\frac{1}{2}}-1)}{x^{\frac{1}{2}}(x^{\frac{1}{2}}-1)}$

$=(x^{\frac{1}{3}}+1)-(1+x^{-\frac{1}{2}})=x^{\frac{1}{3}}-x^{-\frac{1}{2}}$

$\therefore {(\frac{x+1}{x^{\frac{2}{3}}-x^{\frac{1}{3}}+1}-\frac{x-1}{x-x^{\frac{1}{2}}})}^{10}=(x^{\frac{1}{3}}-x^{\frac{-1}{2}}{)}^{10}$

Let $T_{r+1}$ be the general term in $(x^{\frac{1}{3}}-x^{\frac{-1}{2}}{)}^{10}$

Then, $T_{r+1}{=}^{10}{C}_r\left(x^{\frac{1}{3}}{)}^{10-r}\right(-1{)}^r(x^{\frac{1}{2}}{)}^r$

For this term to be independent of x, we must have $\frac{10-r}{3}-\frac{r}{2}=0$

$\Rightarrow 20-2r-3r=0\Rightarrow 5r=20\Rightarrow r=4$

So, required coefficient = ${}^{10}{C}_4(-1{)}^4=210$