The coefficients of three consecutive terms in the expansion of $(1+a)^{n}$ are in the ratio $1: 7: 42 .$ Find $n$
The coefficients of three consecutive terms in the expansion of $(1+a)^{n}$ are in the ratio $1: 7: 42 .$ Find $n$
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, $n=25$
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