The coefficients of three consecutive terms i | vedclass

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(A) Let the three consecutive terms in the expansion of $(1+a)^{n}$ be the $(r-1)^{	ext{th}}$,$r^{	ext{th}}$,and $(r+1)^{	ext{th}}$ terms.The coeff

Vedclass · Accepted Answer

$55$

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(A) Let the three consecutive terms in the expansion of $(1+a)^{n}$ be the $(r-1)^{	ext{th}}$,$r^{	ext{th}}$,and $(r+1)^{	ext{th}}$ terms.The coefficients of these terms are $^{n}C_{r-2}$,$^{n}C_{r-1}$,and $^{n}C_{r}$ respectively.Given the ratio $1:7:42$,we have:$\frac{^{n}C_{r-2}}{^{n}C_{r-1}} = \frac{1}{7} \implies \frac{r-1}{n-r+2} = \frac{1}{7} \implies 7r - 7 = n - r + 2 \implies n - 8r = -9$  ...........$(1)$$\frac{^{n}C_{r-1}}{^{n}C_{r}} = \frac{7}{42} = \frac{1}{6} \implies \frac{r}{n-r+1} = \frac{1}{6} \implies 6r = n - r + 1 \implies n - 7r = -1$  ...........$(2)$Subtracting equation $(1)$ from $(2)$:$(n - 7r) - (n - 8r) = -1 - (-9)$$r = 8$Substituting $r=8$ into equation $(2)$:$n - 7(8) = -1$$n - 56 = -1$$n = 55$

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

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