 QUESTION

# What is the coefficient of ${x^n}$ in the expansion of $\dfrac{1}{{(1 - x)(1 - 2x)(1 - 3x)}}$.(a). $\dfrac{1}{2}({2^{n + 2}} - {3^{n + 3}} + 1)$(b). $\dfrac{1}{2}({3^{n + 2}} - {2^{n + 3}} + 1)$(c). $\dfrac{1}{2}({2^{n + 3}} - {3^{n + 2}} + 1)$(d). None of these

Hint: Use the binomial expansion ${(1 - x)^{ - 1}} = 1 + x + {x^2} + .....$ to simplify the terms in the denominator and then find the coefficients of ${x^n}$ in the expansion. Use the sum of geometric series $\dfrac{{a(1 - {r^n})}}{{1 - r}}$ to simplify the expression.

We need to find the coefficient of ${x^n}$ in the expansion of $\dfrac{1}{{(1 - x)(1 - 2x)(1 - 3x)}}$.
We know that $\dfrac{1}{{(1 - x)(1 - 2x)(1 - 3x)}}$ can be written as ${(1 - x)^{ - 1}}{(1 - 2x)^{ - 1}}{(1 - 3x)^{ - 1}}$.
The formula for binomial expansion for negative exponents is given as follows:
${(1 - x)^{ - 1}} = 1 + x + {x^2} + .....$
Using this formula, we have:
$\dfrac{1}{{(1 - x)(1 - 2x)(1 - 3x)}} = (1 + x + ... + {x^n} + ...)(1 + 2x + ... + {(2x)^n} + ...)(1 + 3x + ... + {(3x)^n} + ...)$
Now, multiplying, we find only the coefficient of ${x^n}$.
Taking 1 from the first bracket and then expressing the multiplication of next two brackets we have:
${C_0} = {3^n} + {2.3^{n - 1}} + .... + {2^{n - 1}}.3 + {2^n}$
Taking the coefficient of x in the first bracket and then expressing the multiplication of next two brackets we have:
${C_1} = {3^{n - 1}} + {2.3^{n - 2}} + .... + {2^{n - 2}}.3 + {2^{n - 1}}$
Similarly, we have until the coefficient of ${x^n}$ in the bracket, which is 1.
${C_n} = 1$
For ${C_0}$, we take ${3^n}$ common and simplify the expression using the geometric sum as $\dfrac{{a(1 - {r^n})}}{{1 - r}}$.
${C_0} = {3^n}\left( {1 + \dfrac{2}{3} + .... + {{\left( {\dfrac{2}{3}} \right)}^{n - 1}} + {{\left( {\dfrac{2}{3}} \right)}^n}} \right)$
${C_0} = {3^n}\left( {\dfrac{{1 - {{\left( {\dfrac{2}{3}} \right)}^{n + 1}}}}{{1 - \dfrac{2}{3}}}} \right)$
Simplifying, we have:
${C_0} = {3^{n + 1}}\left( {1 - {{\left( {\dfrac{2}{3}} \right)}^{n + 1}}} \right)$
${C_0} = {3^{n + 1}} - {2^{n + 1}}$
Similarly, for the ${C_1}$ term, we have:
${C_1} = {3^{n - 1}}\left( {1 + \dfrac{2}{3} + .... + {{\left( {\dfrac{2}{3}} \right)}^{n - 2}} + {{\left( {\dfrac{2}{3}} \right)}^{n - 1}}} \right)$
${C_1} = {3^{n - 1}}\left( {\dfrac{{1 - {{\left( {\dfrac{2}{3}} \right)}^n}}}{{1 - \dfrac{2}{3}}}} \right)$
Simplifying, we have:
${C_1} = {3^n}\left( {1 - {{\left( {\dfrac{2}{3}} \right)}^n}} \right)$
${C_1} = {3^n} - {2^n}$
Adding all the ${C_i}$ terms we have:
$\sum\limits_{i = 0}^n {{C_i} = {3^{n + 1}} - {2^{n + 1}}} + {3^n} - {2^n} + ...... + 3 - 2$
Grouping terms together, we have:
$C = (3 + {3^2} + ........ + {3^n} + {3^{n + 1}}) - (2 + {2^2} + ........ + {2^n} + {2^{n + 1}})$
Using the sum of geometric terms, we have:
$C = \dfrac{{3(1 - {3^{n + 1}})}}{{1 - 3}} - \dfrac{{2(1 - {2^{n + 1}})}}{{1 - 2}}$
Simplifying, we get:
$C = \dfrac{{3({3^{n + 1}} - 1)}}{2} + 2(1 - {2^{n + 1}})$
$C = \dfrac{1}{2}\left[ {{3^{n + 2}} - 3 + 4 - {2^{n + 3}}} \right]$
$C = \dfrac{1}{2}\left[ {{3^{n + 2}} - {2^{n + 3}} + 1} \right]$
Hence, the correct answer is option (b).

Note: You might make a mistake when evaluating the sum of geometric terms. The sum for n terms is $\dfrac{{a(1 - {r^n})}}{{1 - r}}$, the expression $1 + \dfrac{2}{3} + .... + {\left( {\dfrac{2}{3}} \right)^{n - 1}} + {\left( {\dfrac{2}{3}} \right)^n}$ has (n+1) terms. Hence, evaluate accordingly.