Recurrence Relation_discrete

Question

Let $f(x)=\frac 1 {(1-x)^2}$,find the coefficients $a_0, a_1, a_2, \ldots$ in the expansion $f(x)=\sum_{k=0}^\infty a_k x^k$

Solution is using theorem

Let $f(x)=\sum_{k=0}^\infty a_k x^k,\quad g(x)=\sum_{k=0}^\infty b_k x^k$

$$\Rightarrow f(x)g(x)=\sum_{k=0}^\infty \left ( \sum_{j=0}^k a_j b_{k-j} \right) x^k$$

And the solution is given as -: $$\frac 1 {(1-x)^2}=\sum_{k=0}^\infty \left ( \sum_{j=0}^k 1 \right )x^k = \sum_{k=0}^\infty (k+1) x^{k+1} $$

Not getting how to get 1in second summation..please explain ! -

Answer 1

You know that

$$\frac1{1-x}=\sum_{k\ge 0}x^k=\sum_{k\ge 0}\color{crimson}1\cdot x^k\;,$$

so that if we let $a_k$ be the coefficient of $x^k$ in this series, then $\color{crimson}{a_k=1}$ for all $k$. Now

$$\begin{align*} \frac1{(1-x)^2}&=\frac1{1-x}\cdot\frac1{1-x}\\ &=\sum_{k\ge 0}\left(\sum_{j=0}^k\color{crimson}{a_ja_{k-j}}\right)x^k\\ &=\sum_{k\ge 0}\left(\sum_{j=0}^k\color{crimson}1\cdot\color{crimson}1\right)x^k\\ &=\sum_{k\ge 0}\left(\sum_{j=0}^k1\right)x^k\\ &=\sum_{k\ge 0}(k+1)x^k\;. \end{align*}$$

Answer 2

$${\frac {1}{1-x}}=\sum _{n=0}^{\infty }x^{n}\quad {\text{ for }}|x|<1\!$$ differntiate it to get $${\frac {1}{(1-x)^{2}}}=\sum _{n=1}^{\infty }nx^{n-1}$$