Question

What is the coefficient of $$x^n$$ in the expansion of $${\displaystyle \frac{1}{(1-x)(1-2x)(1-3x)}}$$.
(a). $${\displaystyle \frac{1}{2}}(2^{n+2}-3^{n+3}+1)$$
(b). $${\displaystyle \frac{1}{2}}(3^{n+2}-2^{n+3}+1)$$
(c). $${\displaystyle \frac{1}{2}}(2^{n+3}-3^{n+2}+1)$$
(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 $${\displaystyle \frac{a(1-r^n)}{1-r}}$$ to simplify the expression.

Complete step-by-step answer:
We need to find the coefficient of $$x^n$$ in the expansion of $${\displaystyle \frac{1}{(1-x)(1-2x)(1-3x)}}$$.
We know that $${\displaystyle \frac{1}{(1-x)(1-2x)(1-3x)}}$$ 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:
$${\displaystyle \frac{1}{(1-x)(1-2x)(1-3x)}}=(1+x+...+x^n+...)(1+2x+...+(2x{)}^n+...)(1+3x+...+(3x{)}^n+...)$$
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 $${\displaystyle \frac{a(1-r^n)}{1-r}}$$.
$$C_0=3^n\left(1+{\displaystyle \frac{2}{3}}+....+{\left({\displaystyle \frac{2}{3}}\right)}^{n-1}+{\left({\displaystyle \frac{2}{3}}\right)}^n\right)$$
$$C_0=3^n\left({\displaystyle \frac{1-{\left({\displaystyle \frac{2}{3}}\right)}^{n+1}}{1-{\displaystyle \frac{2}{3}}}}\right)$$
Simplifying, we have:
$$C_0=3^{n+1}\left(1-{\left({\displaystyle \frac{2}{3}}\right)}^{n+1}\right)$$
$$C_0=3^{n+1}-2^{n+1}$$
Similarly, for the $$C_1$$ term, we have:
$$C_1=3^{n-1}\left(1+{\displaystyle \frac{2}{3}}+....+{\left({\displaystyle \frac{2}{3}}\right)}^{n-2}+{\left({\displaystyle \frac{2}{3}}\right)}^{n-1}\right)$$
$$C_1=3^{n-1}\left({\displaystyle \frac{1-{\left({\displaystyle \frac{2}{3}}\right)}^n}{1-{\displaystyle \frac{2}{3}}}}\right)$$
Simplifying, we have:
$$C_1=3^n\left(1-{\left({\displaystyle \frac{2}{3}}\right)}^n\right)$$
$$C_1=3^n-2^n$$
Adding all the $$C_i$$ terms we have:
$$\sum _{i=0}^n{C}_i=3^{n+1}-2^{n+1}+3^n-2^n+......+3-2$$
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:
$$C={\displaystyle \frac{3(1-3^{n+1})}{1-3}}-{\displaystyle \frac{2(1-2^{n+1})}{1-2}}$$
Simplifying, we get:
$$C={\displaystyle \frac{3(3^{n+1}-1)}{2}}+2(1-2^{n+1})$$
$$C={\displaystyle \frac{1}{2}}\left[3^{n+2}-3+4-2^{n+3}\right]$$
$$C={\displaystyle \frac{1}{2}}\left[3^{n+2}-2^{n+3}+1\right]$$
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 $${\displaystyle \frac{a(1-r^n)}{1-r}}$$, the expression $$1+{\displaystyle \frac{2}{3}}+....+{\left({\displaystyle \frac{2}{3}}\right)}^{n-1}+{\left({\displaystyle \frac{2}{3}}\right)}^n$$ has (n+1) terms. Hence, evaluate accordingly.