If the coefficients of three consecutive terms in the expansion of ${(1 + x)}^{n}$ are in the ratio $1:7:42$, then the value of $n$ is

Question

Dr. Abubacker · Accepted Answer

(c)
Let $(r + 1)$th, $(r + 2)$th and $(r + 3)$th be three consecutive terms.
Then, $\ ^{n}C_{r}:\ ^{n}C_{r + 1}:\ ^{n}C_{r + 2} = 1:7:42$
Now, $\frac{\ ^{n}C_{r}}{\ ^{n}C_{r + 1}} = \frac{1}{7}\  \Rightarrow \ \ \frac{r + 1}{n - r} = \frac{1}{7}\ \  \Rightarrow \ \ n - 8r = 7$
and $\frac{\ ^{n}C_{r + 1}}{\ ^{n}C_{r + 2}} = \frac{7}{42}$
$\Rightarrow \ \frac{r + 2}{n - r - 1} = \frac{1}{6}$
$\Rightarrow n - 7r = 13$ …(ii)
On solving Eqs. (i) and (ii), we get $n = 55$

Q. If the coefficients of three consecutive terms in the expansion of ${(1 + x)}^{n}$ are in the ratio $1:7:42$, then the value of $n$ is

Explanation

Reviewed by Dr. Abubacker

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