यदि $(1+a)^{n}$ के प्रसार में तीन क्रमागत पदों के गुणांक $1: 7: 42$ के अनुपात में हैं तो $n$ का मान ज्ञात कीजिए।

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, $n=25$