Coefficients of three consecutive terms in expansion of (1+a)^{n} are in ratio 1: 7: 42 . Find n

Hindi

7.Binomial Theorem

hard

The coefficients of three consecutive terms in the expansion of $(1+a)^{n}$ are in the ratio $1: 7: 42 .$ Find $n$

$25$

Solution

Suppose the three consecutive terms in the expansion of $(1+a)^{n}$ are $(r-1)^{ th }, r^{ th }$ and $(r+1)^{ th }$ terms.

The $(r-1)^{\text {th }}$ term is $^{n} C_{r-2} a^{r-2},$ and its coefficient is $^n{C_{r – 2}}.$ Similarly, the coefficients of $r^{\text {th }}$ and $(r+1)^{\text {th }}$ terms are ${\,^n}{C_{r – 1}}$ and $^{n} C_{r},$ respectively.

Since the coefficients are in the ratio $1: 7: 42,$ so we have,

$\frac{{^n{C_{r – 2}}}}{{{\,^n}{C_{r – 1}}}} = \frac{1}{7},$ i.e., $n – 8r + 9 = 0$ ………..$(1)$

and $\frac{{{\,^n}{C_{r – 1}}}}{{{\,^n}{C_r}}} = \frac{7}{{42}},$ i.e., $n – 7r + 1 = 0$ ………..$(2)$

Solving equations $(1)$ and $(2),$ we get, $n=25$

Standard 11

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