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Question: The value of \[C_1^2 + C_2^2 + ... + C_n^2\] (where \[{C_i}\] is the \[{i^{th}}\] coefficient of \[{...

The value of C12+C22+...+Cn2C_1^2 + C_2^2 + ... + C_n^2 (where Ci{C_i} is the ith{i^{th}} coefficient of (1+x)n{(1 + x)^n} expansion) is:

Explanation

Solution

The given equation is based on a special case of the binomial theorem. We can simply expand the equation using the binomial theorem, then square on both the sides and arrive at the answer using the formula: (1+x)n=∑nr−0nCrxr=C0+C1+C2x2+...Cnxn{(1 + x)^n} = \sum\limits_{}^n {_{r - 0}n{C_r}{x^r}} = {C_0} + {C_1} + {C_2}{x^2} + ...{C_n}{x_n} . Finally, we need to find the value of C12+C22+...+Cn2C_1^2 + C_2^2 + ... + C_n^2 by the above formula.

Complete step by step answer:
The Binomial Theorem is a technique for extending an expression elevated to some finite power. A binomial Theorem is a useful expansion method that can be used in Algebra, probability, and other fields. A binomial expression is an algebraic expression which contains two dissimilar terms. Example- a+b,a3+b3a + b,{a^3} + {b^3} etc.

Binomial Theorem can be explained as-
If n∈N,x,y∈Rn \in N,x,y \in R then (x+y)n=n∑r=0nCrxn−ryr{(x + y)^n}{ = ^n}{\sum _{r = 0}}n{C_r}{x^{n - r}}{y^r}
where nCr=n!(n−r)!r!n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}
Points to be noted are:
-The total number of terms in the expansion of (x+y)n{(x + y)^n} are (n+1)(n + 1).
-The sum of exponents of xx and yy is always nn.
-nC0,nC1,nC2,...,nCnn{C_0},n{C_1},n{C_2},...,n{C_n} are called binomial coefficients and also represented by C0,C1,C2,...,Cn{C_0},{C_1},{C_2},...,{C_n}.
-The binomial coefficients which are equidistant from the beginning and from the ending are equal i.e. nC0=nCn,nC1=nCn−1,nC2=nCn−2n{C_0} = n{C_n},n{C_1} = n{C_{n - 1}},n{C_2} = n{C_{n - 2}} etc.

We can expand the equation with the help of formula as follows:
(1+x)n=∑nr=0nCrxr=C0+C1+C2x2+...Cnxn{(1 + x)^n} = \sum\limits_{}^n {_{r = 0}n{C_r}{x^r}} = {C_0} + {C_1} + {C_2}{x^2} + ...{C_n}{x_n}
⇒(1+x)n=nC0+x(nC1)+x2(nC2)+...+xn(nCn)\Rightarrow {(1 + x)^n} = n{C_0} + x(n{C_1}) + {x^2}(n{C_2}) + ... + {x_n}(n{C_n})
Now if we square the equation, on the left-hand side we will get (1+x)2n{(1 + x)^{2n}}.
The coefficient of xn{x^n} in the equation (1+x)2n=2nCn{(1 + x)^{2n}}{ = ^{2n}}{C_n}
Hence, we will get the squared equation as follows:
2nCn=(nC0)2+(nC1)2+(nC2)2+...+(nCn)2^{2n}{C_n} = {(n{C_0})^2} + {(n{C_1})^2} + {(n{C_2})^2} + ... + {(n{C_n})^2}
Where Ci{C_i} is the ith{i^{th}} coefficient of (1+x)n{(1 + x)^n} expansion.

Therefore, we can conclude that: C12+C22+...+Cn2=2nCn=2n!n!n!C_1^2 + C_2^2 + ... + C_n^2{ = ^{2n}}{C_n} = \dfrac{{2n!}}{{n!n!}}.

Note: Here we have assumed nn to be a rational number and xx be a real number such that ∣x∣<1\left| x \right| < 1. To find binomial coefficients we can also use Pascal’s Triangle. Binomial coefficients refer to the integers which are coefficients in the binomial theorem.
C12+C22+...+Cn2=2nCn=2n!n!n!C_1^2 + C_2^2 + ... + C_n^2{ = ^{2n}}{C_n} = \dfrac{{2n!}}{{n!n!}} is one of the most important properties of binomial coefficient.