Problem : Find the sum of the coefficients of the polynomial$ p(x) =(3x-2)^{17}(x+1)^4$
Solution :
$ p(x) =(3x-2)^{17}(x+1)^4 $
$= (a_0+a_1x+….a_{17}x^{17})(b_0+b_1x+….b_4x^4)$ for some $a_i; b_j$
= $(c_0+c_1x+…c_{21}x^{21})$ for some $c_k$
Sum of the coefficients of p(x) =$c_0+c_1+c_2+….c_{21}$
In solution it is given that the sum of the coefficients can be obtained by putting x =1 , my question is why x =1?
Please explain this concept.. else is clear in this problems… Thanks..
Best Answer
If $P(x) = a_{n}x^{n} + a_{n-1}x^{n-1} + \cdots + a_{0}$, and we want to find $a_{n} + a_{n-1} + \cdots + a_{1} + a_{0}$, then we see that:
$P(1) = a_{n}(1)^{n} + a_{n-1}(1)^{n-1} + \cdots + a_{1}(1) + a_{0} = a_{n}(1) + a_{n-1}(1) + \cdots + a_{1}(1) + a_{0} = a_{n} + a_{n-1} + \cdots + a_{1} + a_{0}$