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.

Asked by Pragya Singh | 1 year ago |  69

##### Solution :-

We know that (r + 1)th term, (Tr+1), in the binomial expansion of (a + b)n is given by

Tr+1 = nCr an-t br

The first three terms of the expansion are given as 729, 7290 and 30375 respectively. Then we have,

T1 = nC0 an-0 b0 = an = 729….. 1

T2 = nC1 an-1 b= nan-1 b = 7290…. 2

T3 = nC2 an-2 b2 = {n (n -1)/2 }an-2 b2

= 30375……3

Dividing 2 by 1 we get

$$\dfrac{na^{n-1}b}{a^{n}} = \dfrac{7290}{729} \\ \dfrac{nb}{a} = 10$$

Dividing 3 by 2 we get

$$\dfrac{n(n-1)a^{n-2}b^{2}}{2na^{n-1}b} = \dfrac{30375}{7290} \\$$

$$\dfrac{(n-1)b}{2a} = \dfrac{30375}{7290} \\ \dfrac{(n-1)b}{a} = \dfrac{30375}{7290} \times 2$$

$$=\dfrac{25}{3} \\ \dfrac{(nb)}{a} -\dfrac{b}{a}$$

$$= \dfrac{25}{3} \\$$

$$10 -\dfrac{b}{a} = \dfrac{25}{3} \\$$

$$\dfrac{b}{a} = 10 – \dfrac{25}{3}$$

$$= \dfrac{5}{3}$$

From 4 and 5 we have

n. $$\dfrac{5}{3}$$ = 10

n = 6

Substituting n = 6 in 1 we get

a6 = 729

a = 3

From 5 we have, $$\dfrac{b}{3}$$$$\dfrac{5}{3}$$

b = 5

Thus a = 3, b = 5 and n = 76

Answered by Abhisek | 1 year ago

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