Solveeit Logo
Login

Question

Question: The sum of the coefficients of the polynomial expansion of \({(1 + x - 3{x^2})^{2163}}\) is A. 1 ...

The sum of the coefficients of the polynomial expansion of (1+x−3x2)2163{(1 + x - 3{x^2})^{2163}} is
A. 1
B. -1
C. 0
D. none of these

Explanation

Solution

The given polynomial can be expanded as (1+x−3x2)2163=p(x)=∑n=04326anxn{(1 + x - 3{x^2})^{2163}} = p(x) = \sum\limits_{n = 0}^{4326} {{a_n}{x^n}} where a0,a1,a2,⋯ ,a4325,a4326\\{ {a_{0,}}{a_1},{a_2}, \cdots ,{a_{4325}},{a_{4326}}\\} are coefficients of the following function. We have to put x such that the expansion is independent of xx.

Complete step-by-step answer:
We are given (1+x−3x2)2163{(1 + x - 3{x^2})^{2163}} and we need to find the sum of all the coefficients in expansion. This is a case of multinomial expansion.
First we will show the general expansion of the given polynomial. This is shown as
(1+x−3x2)2163=p(x)=∑n=04326anxn{(1 + x - 3{x^2})^{2163}} = p(x) = \sum\limits_{n = 0}^{4326} {{a_n}{x^n}} … (1)
∑n=04326anxn=a0x0+a1x1+a2x2+................+a4326x4326\sum\limits_{n = 0}^{4326} {{a_n}{x^n}} = {a_0}{x^0} + {a_1}{x^1} + {a_2}{x^2} + ................ + {a_{4326}}{x^{4326}} … (2)
Now, we need to find the value of a0+a1+a2+⋯a4325+a4326{a_0} + {a_1} + {a_2} + \cdots {a_{4325}} + {a_{4326}}. This can be shown as follows
S=∑n=04326anS = \sum\limits_{n = 0}^{4326} {{a_n}}
If we have to find the submission of the coefficients then we must take the value of xx such that the equation must become independent of xx and we get the equation corresponding toa0+a1+a2+⋯a4325+a4326{a_0} + {a_1} + {a_2} + \cdots {a_{4325}} + {a_{4326}}.
Thus, only for x=1x = 1. Thus putting x=1x = 1 in equation (2), we get,
∑n=04326an(1)n=a010+a111+a212+................+a432614326\sum\limits_{n = 0}^{4326} {{a_n}{{(1)}^n}} = {a_0}{1^0} + {a_1}{1^1} + {a_2}{1^2} + ................ + {a_{4326}}{1^{4326}}
Now, putting x=1x = 1in equation (1) we get,
[1+1−3(1)2]2163=∑n=04326an(1)n=∑n=04326an{[1 + 1 - 3{(1)^2}]^{2163}} = \sum\limits_{n = 0}^{4326} {{a_n}{{(1)}^n}} = \sum\limits_{n = 0}^{4326} {{a_n}}
Simplifying the above equation, we get,
∑n=04326an=(−1)2163\sum\limits_{n = 0}^{4326} {{a_n} = } {( - 1)^{2163}}
Since, we know -1 raises to power a negative number, we get -1. Therefore,
∑n=04326an=(−1)2163=−1\sum\limits_{n = 0}^{4326} {{a_n} = } {( - 1)^{2163}} = - 1 ... (3)
Since, equation (2) and equation (3) are similar. We can say that,
a0+a1+a2+⋯a4325+a4326=−1{a_0} + {a_1} + {a_2} + \cdots {a_{4325}} + {a_{4326}} = - 1

Therefore, option (b) -1 is the correct option.

Note: Important points to remember while solving the questions on binomial expansion are:
1. The total number of terms in the expansion of (x+y)n  {\left( {x + y} \right)^n}\; are (n+1).
2. The sum of exponents of x and y is always n.
While these properties are not true for multinomial expansion generally.