Q1Written answer
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.
Show solutionHide solution
Worked 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 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
\(\begin{array}{l}\frac{na^{n-1}b}{a^{n}} = \frac{7290}{729} \\ \frac{nb}{a} = 10\end{array} \)
Dividing 3 by 2, we get
\(\begin{array}{l}\frac{n(n-1)a^{n-2}b^{2}}{2na^{n-1}b} = \frac{30375}{7290} \\ \frac{(n-1)b}{2a} = \frac{30375}{7290} \\ \frac{(n-1)b}{a} = \frac{30375}{7290} \times 2 =\frac{25}{3} \\ \frac{(nb)}{a} -\frac{b}{a} = \frac{25}{3} \\ 10 -\frac{b}{a} = \frac{25}{3} \\ \frac{b}{a} = 10 – \frac{25}{3} = \frac{5}{3}\end{array} \)
From 4 and 5, we have
n. 5/3 = 10
n = 6
Substituting n = 6 in 1, we get
a6 = 729
a = 3
From 5, we have b/3 = 5/3
b = 5
Thus, a = 3, b = 5 and n = 76
is √6: 1
