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Prove that the coefficient of${x^n}$ in the expression of ${(1 + x)^{2n}}$ is twice of the coefficient of the ${x^n}$ in the expression of ${(1 + x)^{2n - 1}}$.

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Last updated date: 13th Jun 2024
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Answer
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Hint: Here we have to compare coefficient of ${x^n}$in both the given expression. For that we will get the expression for (r+1)th term of both ${(1 + x)^{2n}}$and ${(1 + x)^{2n - 1}}$. And from there we will get the coefficient of ${x^n}$.

Complete step-by-step answer:
Step-1
We know that the general form of ${(a + b)^n}$ is
${T_{r + 1}} = {}^n{C_r}{a^{n - r}}{b^r}$
Where, ${T_{r + 1}}$=(r+1)th term
n = highest power of the expression
step-2
In the expression ${(1 + x)^{2n}}$, a=1, b=x, n=2n
The general form of it is,
${T_{r + 1}} = {}^{2n}{C_r}{1^{2n - r}}{x^r}$
$ \Rightarrow {T_{r + 1}} = {}^{2n}{C_r}{x^r}$……………(1)
Step-3
For coefficient of ${x^n}$, putting r=n in equation (1) we get,
${T_{n + 1}} = {}^{2n}{C_n}{x^n}$…………….(2)
Step-4
The coefficient of ${x^n}$ is${}^{2n}{C_n}$
In the expression ${(1 + x)^{2n - 1}}$, a=1, b=x, n=2n-1
The general form of it is,
${T_{r + 1}} = {}^{2n - 1}{C_r}{1^{2n - 1 - r}}{x^r}$
$ \Rightarrow {T_{r + 1}} = {}^{2n - 1}{C_r}{x^r}$……………(3)
Step-5
For coefficient of ${x^n}$, putting r=n in equation (3) we get,
${T_{n + 1}} = {}^{2n - 1}{C_n}{x^n}$……………(4)
The coefficient of ${x^n}$ is ${}^{2n - 1}{C_n}$
Step-6
We are asked to prove that the coefficient of ${x^n}$ in the expression of ${(1 + x)^{2n}}$ is twice of the coefficient of the ${x^n}$ in the expression of ${(1 + x)^{2n - 1}}$
i.e. ${}^{2n}{C_n}$= 2${}^{2n - 1}{C_n}$
step-7
Simplifying left hand side,
${}^{2n}{C_n}$
$ = \dfrac{{2n!}}{{n!(2n - n)!}}$
$ = \dfrac{{2n!}}{{n!n!}}$
Step-8
Again simplifying right hand side we get,
2${}^{2n - 1}{C_n}$
$ = 2 \times \dfrac{{(2n - 1)!}}{{n!(2n - 1 - n)!}}$
$ = 2 \times \dfrac{{(2n - 1)!}}{{n!(n - 1)!}}$
Step-9
Multiplying and dividing by n we get,
$ = 2 \times \dfrac{{(2n - 1)!}}{{n!(n - 1)!}} \times \dfrac{n}{n}$
$ = \dfrac{{2n(2n - 1)!}}{{n!n(n - 1)!}}$
$ = \dfrac{{2n!}}{{n!n!}}$
Step-10
It is proved that L.H.S = R.H.S
Hence it is proved that the coefficient of ${x^n}$ in the expression of ${(1 + x)^{2n}}$ is twice of the coefficient of the ${x^n}$ in the expression of ${(1 + x)^{2n - 1}}$.

Note: This is a binomial sequence and series question. Be cautious while solving expansion of general form of expression and while comparing them.
Binomial sequences can be used to prove results and solve problems in combinatorics, algebra, calculus, and many other areas of mathematics.