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Question: The coefficient of \(x^{53}\) in \( \sum\limits_{m = 0}^{100} {{}^{100}{C_m}} {\left( {x - 3} \right...

The coefficient of x53x^{53} in ∑m=0100100Cm(x−3)100−m2m \sum\limits_{m = 0}^{100} {{}^{100}{C_m}} {\left( {x - 3} \right)^{100 - m}}{2^m} is:

  1. 100C47{}^{100}{C_{47}}
  2. 100C53{}^{100}{C_{53}}
  3. −100C53{}^{-100}{C_{53}}
  4. −100C100{}^{-100}{C_{100}}
Explanation

Solution

The above problem is based on the binomial expansion which is used when an expression or an exponent is raised to higher powers.
The general term of the Binomial expansion is:
⇒\Rightarrow (x+y)n=nC0xn+nC1xn−1y+nC2xn−2y2+...........xCnyx{\left( {x + y} \right)^n} = {}^n{C_0}{x^n} + {}^n{C_1}{x^{n - 1}}y + {}^n{C_2}{x^{n - 2}}{y^2} + {...........^x}{C_n}{y^x}
The above expression can be written as;
⇒\Rightarrow (x+y)n=∑0nnCrxn−ryr{\left( {x + y} \right)^n} = \sum\limits_0^n {{}^n{C_r}} {x^{n - r}}{y^r}
Where x and y are variables, n is the power to which an expression is raised and C is the coefficient of combination.
Using the above relations we will solve the given problem.

Complete step-by-step answer:
In order to solve the above problem first we will understand the Binomial theorem in detail and then we will proceed for the calculation part of the problem.
The Binomial theorem describes the algebraic expansion of powers of a binomial.
According to the Binomial theorem, it is possible to expand the polynomial (x + y)n into a sum involving the terms of the form a xb yc, where the exponents b and c are nonnegative integers with b + c =n, and the coefficient a of each term is a specific positive integer depending on n and b.
Now we will perform the calculation part;
As per the equation given to us:
⇒∑m=0100100Cm(x−3)100−m2m\Rightarrow \sum\limits_{m = 0}^{100} {{}^{100}{C_m}} {\left( {x - 3} \right)^{100 - m}}{2^m}
⇒[(x−3)+2]100\Rightarrow {[(x - 3) + 2]^{100}}
⇒[x−1]100\Rightarrow {[x - 1]^{100}}
The above expression can be written as
⇒[1−x]100\Rightarrow {[1 - x]^{100}} (power is even therefore we change the sign inside the bracket)...........1
We can expand the equation 1 as per the below generalized equation;
⇒\Rightarrow (1−x)n=nC0−nC1x+nC2x2......................(−1)rCrxr{\left( {1 - x} \right)^n} = {}^n{C_0} - {}^n{C_1}x + {}^n{C_2}{x^2}......................{( - 1)^{r}}{C_r}{x^r}
⇒(1−x)100=100C0x100−100C1x99+100C2x98......................100C100x100\Rightarrow {\left( {1 - x} \right)^{100}} = {}^{100}{C_0}{x^{100}} - {}^{100}{C_1}{x^{99}} + {}^{100}{C_2}{x^{98}}......................{}^{100}{C_{100}}{x^{100}}
In the above series when power of x becomes 53, it is odd power which will make the term negative.
Therefore, the term will become 100C53(−x)53{}^{100}{C_{53}}{\left( { - x} \right)^{53}} or we can write −100C53{}^{ - 100}{C_{53}}.

Option 3 is correct.

Note: Binomial theorem and its generalization can be used to prove results and solve problems in combinations, algebra, calculus and many areas of mathematics. Binomial theorem helps in exploring probability in an organized way and statistical calculations as well.