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Question: What is the coefficient of \[{x^n}\] in the expansion of \[\dfrac{1}{{(1 - x)(1 - 2x)(1 - 3x)}}\]. ...

What is the coefficient of xn{x^n} in the expansion of 1(1−x)(1−2x)(1−3x)\dfrac{1}{{(1 - x)(1 - 2x)(1 - 3x)}}.
(a). 12(2n+2−3n+3+1)\dfrac{1}{2}({2^{n + 2}} - {3^{n + 3}} + 1)
(b). 12(3n+2−2n+3+1)\dfrac{1}{2}({3^{n + 2}} - {2^{n + 3}} + 1)
(c). 12(2n+3−3n+2+1)\dfrac{1}{2}({2^{n + 3}} - {3^{n + 2}} + 1)
(d). None of these

Explanation

Solution

Hint: Use the binomial expansion (1−x)−1=1+x+x2+.....{(1 - x)^{ - 1}} = 1 + x + {x^2} + ..... to simplify the terms in the denominator and then find the coefficients of xn{x^n} in the expansion. Use the sum of geometric series a(1−rn)1−r\dfrac{{a(1 - {r^n})}}{{1 - r}} to simplify the expression.

Complete step-by-step answer:
We need to find the coefficient of xn{x^n} in the expansion of 1(1−x)(1−2x)(1−3x)\dfrac{1}{{(1 - x)(1 - 2x)(1 - 3x)}}.
We know that 1(1−x)(1−2x)(1−3x)\dfrac{1}{{(1 - x)(1 - 2x)(1 - 3x)}} can be written as (1−x)−1(1−2x)−1(1−3x)−1{(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+x2+.....{(1 - x)^{ - 1}} = 1 + x + {x^2} + .....
Using this formula, we have:
1(1−x)(1−2x)(1−3x)=(1+x+...+xn+...)(1+2x+...+(2x)n+...)(1+3x+...+(3x)n+...)\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 xn{x^n}.
Taking 1 from the first bracket and then expressing the multiplication of next two brackets we have:
C0=3n+2.3n−1+....+2n−1.3+2n{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:
C1=3n−1+2.3n−2+....+2n−2.3+2n−1{C_1} = {3^{n - 1}} + {2.3^{n - 2}} + .... + {2^{n - 2}}.3 + {2^{n - 1}}
Similarly, we have until the coefficient of xn{x^n} in the bracket, which is 1.
Cn=1{C_n} = 1
For C0{C_0}, we take 3n{3^n} common and simplify the expression using the geometric sum as a(1−rn)1−r\dfrac{{a(1 - {r^n})}}{{1 - r}}.
C0=3n(1+23+....+(23)n−1+(23)n){C_0} = {3^n}\left( {1 + \dfrac{2}{3} + .... + {{\left( {\dfrac{2}{3}} \right)}^{n - 1}} + {{\left( {\dfrac{2}{3}} \right)}^n}} \right)
C0=3n(1−(23)n+11−23){C_0} = {3^n}\left( {\dfrac{{1 - {{\left( {\dfrac{2}{3}} \right)}^{n + 1}}}}{{1 - \dfrac{2}{3}}}} \right)
Simplifying, we have:
C0=3n+1(1−(23)n+1){C_0} = {3^{n + 1}}\left( {1 - {{\left( {\dfrac{2}{3}} \right)}^{n + 1}}} \right)
C0=3n+1−2n+1{C_0} = {3^{n + 1}} - {2^{n + 1}}
Similarly, for the C1{C_1} term, we have:
C1=3n−1(1+23+....+(23)n−2+(23)n−1){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)
C1=3n−1(1−(23)n1−23){C_1} = {3^{n - 1}}\left( {\dfrac{{1 - {{\left( {\dfrac{2}{3}} \right)}^n}}}{{1 - \dfrac{2}{3}}}} \right)
Simplifying, we have:
C1=3n(1−(23)n){C_1} = {3^n}\left( {1 - {{\left( {\dfrac{2}{3}} \right)}^n}} \right)
C1=3n−2n{C_1} = {3^n} - {2^n}
Adding all the Ci{C_i} terms we have:
∑i=0nCi=3n+1−2n+1+3n−2n+......+3−2\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+32+........+3n+3n+1)−(2+22+........+2n+2n+1)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=3(1−3n+1)1−3−2(1−2n+1)1−2C = \dfrac{{3(1 - {3^{n + 1}})}}{{1 - 3}} - \dfrac{{2(1 - {2^{n + 1}})}}{{1 - 2}}
Simplifying, we get:
C=3(3n+1−1)2+2(1−2n+1)C = \dfrac{{3({3^{n + 1}} - 1)}}{2} + 2(1 - {2^{n + 1}})
C=12[3n+2−3+4−2n+3]C = \dfrac{1}{2}\left[ {{3^{n + 2}} - 3 + 4 - {2^{n + 3}}} \right]
C=12[3n+2−2n+3+1]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 a(1−rn)1−r\dfrac{{a(1 - {r^n})}}{{1 - r}}, the expression 1+23+....+(23)n−1+(23)n1 + \dfrac{2}{3} + .... + {\left( {\dfrac{2}{3}} \right)^{n - 1}} + {\left( {\dfrac{2}{3}} \right)^n} has (n+1) terms. Hence, evaluate accordingly.