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

Answer 1

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.

Complete step-by-step answer:
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.