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Question: Prove that the coefficient of\({x^n}\) in the expression of \({(1 + x)^{2n}}\) is twice of the coeff...

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

Explanation

Solution

Here we have to compare coefficient of xn{x^n}in both the given expression. For that we will get the expression for (r+1)th term of both (1+x)2n{(1 + x)^{2n}}and (1+x)2n−1{(1 + x)^{2n - 1}}. And from there we will get the coefficient of xn{x^n}.

Complete step-by-step answer:
Step-1
We know that the general form of (a+b)n{(a + b)^n} is
Tr+1=nCran−rbr{T_{r + 1}} = {}^n{C_r}{a^{n - r}}{b^r}
Where, Tr+1{T_{r + 1}}=(r+1)th term
n = highest power of the expression
step-2
In the expression (1+x)2n{(1 + x)^{2n}}, a=1, b=x, n=2n
The general form of it is,
Tr+1=2nCr12n−rxr{T_{r + 1}} = {}^{2n}{C_r}{1^{2n - r}}{x^r}
⇒Tr+1=2nCrxr\Rightarrow {T_{r + 1}} = {}^{2n}{C_r}{x^r}……………(1)
Step-3
For coefficient of xn{x^n}, putting r=n in equation (1) we get,
Tn+1=2nCnxn{T_{n + 1}} = {}^{2n}{C_n}{x^n}…………….(2)
Step-4
The coefficient of xn{x^n} is2nCn{}^{2n}{C_n}
In the expression (1+x)2n−1{(1 + x)^{2n - 1}}, a=1, b=x, n=2n-1
The general form of it is,
Tr+1=2n−1Cr12n−1−rxr{T_{r + 1}} = {}^{2n - 1}{C_r}{1^{2n - 1 - r}}{x^r}
⇒Tr+1=2n−1Crxr\Rightarrow {T_{r + 1}} = {}^{2n - 1}{C_r}{x^r}……………(3)
Step-5
For coefficient of xn{x^n}, putting r=n in equation (3) we get,
Tn+1=2n−1Cnxn{T_{n + 1}} = {}^{2n - 1}{C_n}{x^n}……………(4)
The coefficient of xn{x^n} is 2n−1Cn{}^{2n - 1}{C_n}
Step-6
We are asked to prove that the coefficient of xn{x^n} in the expression of (1+x)2n{(1 + x)^{2n}} is twice of the coefficient of the xn{x^n} in the expression of (1+x)2n−1{(1 + x)^{2n - 1}}
i.e. 2nCn{}^{2n}{C_n}= 22n−1Cn{}^{2n - 1}{C_n}
step-7
Simplifying left hand side,
2nCn{}^{2n}{C_n}
=2n!n!(2n−n)!= \dfrac{{2n!}}{{n!(2n - n)!}}
=2n!n!n!= \dfrac{{2n!}}{{n!n!}}
Step-8
Again simplifying right hand side we get,
22n−1Cn{}^{2n - 1}{C_n}
=2×(2n−1)!n!(2n−1−n)!= 2 \times \dfrac{{(2n - 1)!}}{{n!(2n - 1 - n)!}}
=2×(2n−1)!n!(n−1)!= 2 \times \dfrac{{(2n - 1)!}}{{n!(n - 1)!}}
Step-9
Multiplying and dividing by n we get,
=2×(2n−1)!n!(n−1)!×nn= 2 \times \dfrac{{(2n - 1)!}}{{n!(n - 1)!}} \times \dfrac{n}{n}
=2n(2n−1)!n!n(n−1)!= \dfrac{{2n(2n - 1)!}}{{n!n(n - 1)!}}
=2n!n!n!= \dfrac{{2n!}}{{n!n!}}
Step-10
It is proved that L.H.S = R.H.S
Hence it is proved that the coefficient of xn{x^n} in the expression of (1+x)2n{(1 + x)^{2n}} is twice of the coefficient of the xn{x^n} in the expression of (1+x)2n−1{(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.