Question

The value of \[C_1^2 + C_2^2 + ... + C_n^2\] (where \[{C_i}\] is the \[{i^{th}}\] coefficient of \[{(1 + x)^n}\] expansion) is:

Answer 1

Hint: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: \[{(1 + x)^n} = \sum\limits_{}^n {_{r - 0}n{C_r}{x^r}} = {C_0} + {C_1} + {C_2}{x^2} + ...{C_n}{x_n}\] . Finally, we need to find the value of \[C_1^2 + C_2^2 + ... + C_n^2\] 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- \[a + b,{a^3} + {b^3}\] etc.

Binomial Theorem can be explained as-
If \[n \in N,x,y \in R\] then \[{(x + y)^n}{ = ^n}{\sum _{r = 0}}n{C_r}{x^{n - r}}{y^r}\]
where \[n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\]
Points to be noted are:
-The total number of terms in the expansion of \[{(x + y)^n}\] are \[(n + 1)\].
-The sum of exponents of \[x\] and \[y\] is always \[n\].
-\[n{C_0},n{C_1},n{C_2},...,n{C_n}\] are called binomial coefficients and also represented by \[{C_0},{C_1},{C_2},...,{C_n}\].
-The binomial coefficients which are equidistant from the beginning and from the ending are equal i.e. \[n{C_0} = n{C_n},n{C_1} = n{C_{n - 1}},n{C_2} = n{C_{n - 2}}\] etc.

We can expand the equation with the help of formula as follows:
\[{(1 + x)^n} = \sum\limits_{}^n {_{r = 0}n{C_r}{x^r}} = {C_0} + {C_1} + {C_2}{x^2} + ...{C_n}{x_n}\]
\[\Rightarrow {(1 + x)^n} = n{C_0} + x(n{C_1}) + {x^2}(n{C_2}) + ... + {x_n}(n{C_n})\]
Now if we square the equation, on the left-hand side we will get \[{(1 + x)^{2n}}\].
The coefficient of \[{x^n}\] in the equation \[{(1 + x)^{2n}}{ = ^{2n}}{C_n}\]
Hence, we will get the squared equation as follows:
\[^{2n}{C_n} = {(n{C_0})^2} + {(n{C_1})^2} + {(n{C_2})^2} + ... + {(n{C_n})^2}\]
Where \[{C_i}\] is the \[{i^{th}}\] coefficient of \[{(1 + x)^n}\] expansion.

Therefore, we can conclude that: \[C_1^2 + C_2^2 + ... + C_n^2{ = ^{2n}}{C_n} = \dfrac{{2n!}}{{n!n!}}\].

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