Question
Let $${\left( {1 + x} \right)^n} = \sum\limits_{r = 0}^n {{a_r}{x^r}.} $$ Then $$\left( {1 + \frac{{{a_1}}}{{{a_0}}}} \right)\left( {1 + \frac{{{a_2}}}{{{a_1}}}} \right).....\left( {1 + \frac{{{a_n}}}{{{a_{n - 1}}}}} \right)$$ is equal to
A.
$$\frac{{{{\left( {n + 1} \right)}^{n + 1}}}}{{n!}}$$
B.
$$\frac{{{{\left( {n + 1} \right)}^n}}}{{n!}}$$
C.
$$\frac{{{n^{n - 1}}}}{{\left( {n - 1} \right)!}}$$
D.
$$\frac{{{{\left( {n + 1} \right)}^{n - 1}}}}{{\left( {n - 1} \right)!}}$$
Answer :
$$\frac{{{{\left( {n + 1} \right)}^n}}}{{n!}}$$
Solution :
Expression $$ = \frac{{^n{C_0} + {\,^n}{C_1}}}{{{\,^n}{C_0}}} \cdot \frac{{{\,^n}{C_1} + {\,^n}{C_2}}}{{{\,^n}{C_1}}}.....\frac{{{\,^n}{C_{n - 1}} + {\,^n}{C_n}}}{{{\,^n}{C_{n - 1}}}}$$
$$\eqalign{
& = \frac{{{\,^{n + 1}}{C_1}}}{{{\,^n}{C_0}}} \cdot \frac{{{\,^{n + 1}}{C_2}}}{{{\,^n}{C_1}}}.....\frac{{{\,^{n + 1}}{C_n}}}{{{\,^n}{C_{n - 1}}}} \cr
& = \frac{{1 \cdot {\,^{n + 1}}{C_1}}}{{{\,^n}{C_0}}} \cdot \frac{{2 \cdot {\,^{n + 1}}{C_2}}}{{{\,^n}{C_1}}}.....\frac{{n \cdot {\,^{n + 1}}{C_n}}}{{{\,^n}{C_{n - 1}}}} \cdot \frac{1}{{n!}} \cr
& = \frac{{\left( {n + 1} \right) \cdot {\,^n}{C_0}}}{{{\,^n}{C_0}}} \cdot \frac{{\left( {n + 1} \right) \cdot {\,^n}{C_1}}}{{{\,^n}{C_1}}}.....\frac{{\left( {n + 1} \right) \cdot {\,^n}{C_{n - 1}}}}{{{\,^n}{C_{n - 1}}}} \cdot \frac{1}{{n!}}. \cr} $$