Question

If $(1+2x+3x^2{)}^{10}=b_0+b_{1}x+b_2{x}^2....+b_{20}{x}^{20}$, then $b_1$ = 20, $b_2$ = 210, $b_4$ = 8085 and $b_{20}$ = $3^{10}$

Answer 1

$(x+yz{)}^n=\sum \frac{n!}{p!q!r!}{x}^p.y^q.z^r.$
where p + q + r = n
$\therefore (1+2x+3x^2{)}^{10}=\sum \frac{10!}{p!q!r!}{1}^p.(2x{)}^q(3x^2{)}^r$
$=\sum \frac{10!}{p!q!r!}{2}^q{.3}^r.x^{q+2r}$
where p + q + r = 10
Now $b_1,b_2,b_4,b_{20}$ are coefficients of $x,x^2,x^4$ and $x^{20}$ respectively. Hence we must have
q + 2r = 1 or 1 or 4 or 20
We will now solve (1) with each of the four equations given in (2) to get the non-negative integral values of p,q and r.
p + q + r = 10, q + 2r = 1
$\therefore$ r = 0, q = 1, p = 9
$\therefore b_1=\frac{10!}{9!1!0!}{1}^9.\left(2{)}^{1}1\right(3{)}^0=\frac{10.9!\times 2}{9!\times 1\times 1}=20$
Again solving for non-negative integral values
p + q + r = 10 and q + 2r = 2
$\therefore$ r = 0, q = 2, p = 8, r = 1 or q = 0, p = 9
$\therefore b_2=\frac{10!}{8!2!0!}(1{)}^8.(2{)}^2.!(3{)}^0+\frac{10!}{9!1!0!}{1}^9.(2{)}^0(3{)}^1$.
$b_2$ = (45) (4) + (10) (3) = 180 + 30 = 210
Again solving p + q + r = 10 and q + 2r = 4
$\therefore$ r = 0, q = 4 p = 6, or r = 1, q = 2, p = 6
or r = 2, q = 0, p = 8
$\therefore b_4=\frac{10!}{6!4!0!}{1}^6{.2}^4{.3}^0+\frac{10!}{6!2!1!}{1}^6{.2}^2{.3}^1+\frac{10!}{8!0!2!}{1}^6{.2}^0{.3}^2$
$b_4$ = 3360 + 4320 + 405 = 8085
Again solving
p + q + r = 10 and q + 2r = 20
We have r = 10, q = 0, p = 0. All other values of r from 0 to 9 and corresponding values of q will make q + r > 10 whereas p + q + r = 10
$\therefore b_{20}=\frac{10!}{10!}{1}^0{.2}^0{3.}^{10}=3^{10}$