Question

If ${\sum }_{r=0}^{2n}{a}_r(x-2{)}^2={\sum }_{i=0}^{2n}{b}_r(x-3{)}^{r}and{a}_k=1$ for all $k\ge n$ , then show $b_n={}^{2n+1}{C}_{n+1}$

Answer 1

Put x - 3 = t that x -2 = t +1
$\therefore {\sum }_{r=0}^n{a}_r(1+t{)}^r={\sum }_{r=0}^n{b}_r{t}^r$
$b_n$ is coefficient of $t_n$ in R . H. S. so that it will be coefficient of $t_n$ in the L.H.S. , But r = 0 to n - 1 no terms we have the term of $t_{n}andforr\ge n,a_r=1$
$\therefore L.H.S.{\sum }_{r=0}^{n}1\cdot (1+t{)}^r$
$=(1+t{)}^n÷(1+t{)}^{n+1}+(1+t{)}^{n+2}+...+(1+t{)}^{n+n}$
Above is a G .P of n + 1 terms common ratio 1 + t
L. H. S. $=(1+r{)}^n\left[\frac{(1+t{)}^{n+1}-1}{(1-t)-1}\right]$
$=(1+t{)}^{2n+1}-(1+t{)}^{n}t$
We have to search a coefficient of $t_n$ in L.H. S. or coefficient of $t^{n+1}$ is numerator of L. H. S. i.e. $(1+t{)}^{2n+1}-(1+t{)}^n.$
It is ${}^{2n+1}{C}_{n}in{T}_{n=2}$ of 1st only as the 2nd term will not have $t_{n+1}$
$\therefore 2n+1C_n=b_n$