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  $$a =  - 1,\;b =  - 2$$ .

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