Question
Question: If \[\left| x \right|<1\], then the coefficient of \[{{x}^{n}}\] in the expansion of \[{{\left( 1+x+...
If ∣x∣<1, then the coefficient of xn in the expansion of (1+x+x2+x3+.....)2 is
A. n
B. n-1
C. n+2
D. n+1
Solution
In this problem, we have to find the coefficient of xn in the expansion of (1+x+x2+x3+.....)2. We should know the binomial theorem formula for negative index to find the coefficient of xn. We can first change the given expression into a binomial theorem to find the coefficient of xn. We can find the coefficient term in a step by step manner.
Complete step by step solution:
We know that the binomial theorem for negative index 1 is
(1−x)−1=1+x+x2+x3+.....+xn
We know that the given expansion is,
(1+x+x2+x3+.....)2
We can see that the above binomial theorem formula is similar to the given expression as it has been squared, so we can square the binomial theorem formula to get the given expression value, we get