Find a, b and n in the expansion of (a+b)^{n} | vedclass

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(A) The $(r+1)^{th}$ term in the binomial expansion of $(a+b)^{n}$ is given by $T_{r+1} = {}^{n}C_{r} a^{n-r} b^{r}$.The first three terms are:$T_{1}

Vedclass · Accepted Answer

$a=3, b=5, n=6$

Vedclass · Answer

(A) The $(r+1)^{th}$ term in the binomial expansion of $(a+b)^{n}$ is given by $T_{r+1} = {}^{n}C_{r} a^{n-r} b^{r}$.The first three terms are:$T_{1} = {}^{n}C_{0} a^{n} b^{0} = a^{n} = 729$ $(1)$$T_{2} = {}^{n}C_{1} a^{n-1} b^{1} = n a^{n-1} b = 7290$ $(2)$$T_{3} = {}^{n}C_{2} a^{n-2} b^{2} = \frac{n(n-1)}{2} a^{n-2} b^{2} = 30375$ $(3)$Dividing $(2)$ by $(1)$:$\frac{n a^{n-1} b}{a^{n}} = \frac{7290}{729} \Rightarrow \frac{n b}{a} = 10$ $(4)$Dividing $(3)$ by $(2)$:$\frac{n(n-1) a^{n-2} b^{2}}{2 n a^{n-1} b} = \frac{30375}{7290}$ $\Rightarrow \frac{(n-1) b}{2 a} = \frac{30375}{7290} = \frac{25}{6}$$\Rightarrow \frac{(n-1) b}{a} = \frac{25}{3}$ $(5)$From $(4)$,$b = \frac{10a}{n}$. Substituting into $(5)$:$\frac{(n-1) 10a}{n a} = \frac{25}{3} \Rightarrow \frac{10(n-1)}{n} = \frac{25}{3}$$30(n-1) = 25n$ $\Rightarrow 30n - 30 = 25n$ $\Rightarrow 5n = 30$ $\Rightarrow n = 6$.Using $n=6$ in $(4)$:$\frac{6b}{a} = 10$ $\Rightarrow \frac{b}{a} = \frac{10}{6} = \frac{5}{3}$ $\Rightarrow b = \frac{5a}{3}$.Using $n=6$ in $(1)$:$a^{6} = 729 = 3^{6} \Rightarrow a = 3$.Then $b = \frac{5(3)}{3} = 5$.Thus,$a=3, b=5, n=6$.

Find $a, b$ and $n$ in the expansion of $(a+b)^{n}$ if the first three terms of the expansion are $729, 7290$ and $30375$ respectively.

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