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 b1 = 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
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