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Question: If \( {{C}_{0}},{{C}_{1}},{{C}_{2}} \) denotes coefficients expansion of \( {{\left( 1+x \right)}^{n...

If C0,C1,C2{{C}_{0}},{{C}_{1}},{{C}_{2}} denotes coefficients expansion of (1+x)n{{\left( 1+x \right)}^{n}} ,then C1+C1C2+C2C3+..........+Cn−1Cn=(2n)!(n−1)!(n+1)!{{C}_{1}}+{{C}_{1}}{{C}_{2}}+{{C}_{2}}{{C}_{3}}+..........+{{C}_{n-1}}{{C}_{n}}=\dfrac{\left( 2n \right)!}{\left( n-1 \right)!\left( n+1 \right)!}

Explanation

Solution

Hint : The binomial theorem states that (a+b)n=C0an+C1an−1b1+C2an−2b2+......+Cna0bn{{\left( a+b \right)}^{n}}={{C}_{0}}{{a}^{n}}+{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+......+{{C}_{n}}{{a}^{0}}{{b}^{n}} where n is a positive integer. Try finding the relation between the coefficients given on the left hand side of the equation.

Complete step-by-step answer :
Binomial theorem is a quick way of expanding a binomial expression that has been raised to some large power.
Whenever a binomial is raised to some power, the coefficients that are the constant terms associated with the variable terms form a pattern.
Each expansion has one more term than the power on the binomial and the sum of the exponents in each term in the expansion is the same as the power of the binomial.
Pascal’s triangle is an alternative way of determining the coefficients in binomial expansion.
According to the BINOMIAL THEOREM, n being a positive integer:
(a+b)n=C0an+C1an−1b1+C2an−2b2+......+Cna0bn{{\left( a+b \right)}^{n}}={{C}_{0}}{{a}^{n}}+{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+......+{{C}_{n}}{{a}^{0}}{{b}^{n}}
Where Cr=n!r!(n−r)!{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}
Where n!n! is called the factorial of a non-negative integer n and is the product of all positive integers less than or equal to nn .
On expansion of (1+x)n{{\left( 1+x \right)}^{n}} , we get
(1+x)n=C0+C1x+C2x2+C3x3+......+Cnxn{{\left( 1+x \right)}^{n}}={{C}_{0}}+{{C}_{1}}x+{{C}_{2}}{{x}^{2}}+{{C}_{3}}{{x}^{3}}+......+{{C}_{n}}{{x}^{n}}
(1+x)n=C0+C1x+C2x2+C3x3+......+Cnxn{{\left( 1+x \right)}^{n}}={{C}_{0}}+{{C}_{1}}x+{{C}_{2}}{{x}^{2}}+{{C}_{3}}{{x}^{3}}+......+{{C}_{n}}{{x}^{n}} --equation 1
It can also be written as,
(1+x)n=Cnxn+Cn−1xn−1+.......+C2x2+C1x+C0{{\left( 1+x \right)}^{n}}={{C}_{n}}{{x}^{n}}+{{C}_{n-1}}{{x}^{n-1}}+.......+{{C}_{2}}{{x}^{2}}+{{C}_{1}}x+{{C}_{0}} --equation 2
Multiplying equation 1 and equation 2
(1+x)2n=(C0+C1x+C2x2+....+Cnxn)(Cnxn+Cn−1xn−1+.....+C1x+C0){{\left( 1+x \right)}^{2n}}=\left( {{C}_{0}}+{{C}_{1}}x+{{C}_{2}}{{x}^{2}}+....+{{C}_{n}}{{x}^{n}} \right)\left( {{C}_{n}}{{x}^{n}}+{{C}_{n-1}}{{x}^{n-1}}+.....+{{C}_{1}}x+{{C}_{0}} \right) --equation 3
There is no need to multiply the whole term.
We observe the left hand side of the question.
We observe that the coefficients for the term xn−1{{x}^{n-1}} are C1+C1C2+C2C3+..........+Cn−1Cn{{C}_{1}}+{{C}_{1}}{{C}_{2}}+{{C}_{2}}{{C}_{3}}+..........+{{C}_{n-1}}{{C}_{n}} .
Thus we need to find the coefficients of xn−1{{x}^{n-1}} on the right hand side of equation 3 as well.
Coefficient of xn−1{{x}^{n-1}} in (1+x)2n{{\left( 1+x \right)}^{2n}} is 2nCn−1^{2n}{{C}_{n-1}}
On evaluating 2nCn−1^{2n}{{C}_{n-1}} we get,
2nCn−1^{2n}{{C}_{n-1}} = (2n)!(2n−(n−1))!(n−1)!\dfrac{\left( 2n \right)!}{\left( 2n-\left( n-1 \right) \right)!\left( n-1 \right)!} = (2n)!(n+1)!(n−1)!\dfrac{\left( 2n \right)!}{\left( n+1 \right)!\left( n-1 \right)!}
Hence Proved!

Note : We can also solve this question by analyzing the right hand side of the question and converting it into the form of nCr^{n}{{C}_{r}} . Then we come to know the value of r and hence can find the coefficient relation of xr{{x}^{r}} .