# Problem Study (The Binomial Theorem)

The second, third and fourth terms in the expansion of (a + b)n are 12, 60 and 160 respectively. Find the values of a, b and n.
$\begin{array}{l} {\left( {a + b} \right)^n} = {}^n{C_0}{a^n} + {}^n{C_1}{a^{n - 1}} b + {}^n{C_2}{a^{n - 2}} {b^2} + {}^n{C_3}{a^{n - 3}} {b^3} + ...\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;= {a^n} + n{a^{n - 1}}b + \frac{{n(n - 1)}}{{1 \times 2}}{a^{n - 2}}{b^2} + \frac{{n(n - 1)(n - 2)}}{{1 \times 2 \times 3}}{a^{n - 3}}{b^3} + ... \end{array}$
By the problem,
$\begin{array}{l} n{a^{n - 1}}b = 12 - - - - - - - - - (1)\\ \frac{{n(n - 1)}}{{1 \times 2}}{a^{n - 2}}{b^2} = 60 - - - - - - (2)\\ \frac{{n(n - 1)(n - 2)}}{{1 \times 2 \times 3}}{a^{n - 3}}{b^3} = 160 - - - (3) \end{array}$
Dividing equation (2) by equation (1),
$\frac{{\frac{{n(n - 1)}}{2}{a^n}^{ - 2}{b^2}}}{{n{a^n}^{ - 1}b}} = 5$
which yields
$\frac{b}{a} = \frac{{10}}{{n - 1}} - - - - - - (4)$
Again, equation (3) is divided by equation (2),
$\frac{{\frac{{n(n - 1)(n - 2)}}{6}{a^n}^{--3}{b^3}}}{{\frac{{n(n - 1)}}{2}{a^n}^{--2}{b^2}}} = \frac{{160}}{{60}}$
yields
$\frac{b}{a} = \frac{8}{{n - 2}} - - - - - - (5)$
Hence by equation (4) and (5),
$\frac{{10}}{{n - 1}} = \frac{8}{{n - 2}}$
Solving the equation, we get    $n = 6$
Substituting n = 6  in equation (4) or (5), we get
$\frac{b}{a} = 2 \Rightarrow b = 2a$
But we have ..$n{a^{n - 1}}b = 12$
$(6){a^{6 - 1}}(2a) = 12$
Hence ${a^6} = 1 \Rightarrow a = \pm 1$
So, when $a = - 1,\;b = - 2$ and when $a = - 1,\;b = - 2$.