Question

State and prove binomial theorem for positive integral index 'n'

Answer 1

What is a Binomial Expression?

An algebraic expression consisting of two different terms is called a binomial expression. Eg: Terms of the form x3 + y3 are binomial expressions.

What is the Binomial Theorem for a positive integral?

The binomial theorem explains the way of expressing and evaluating the powers of a binomial. This theorem explains that a term of the form (a+b)n can be expanded and expressed in the form of rasbt, where the exponents s and t are non-negative integers satisfying the condition s + t = n. The coefficient r is a positive integer. The terms involved are called binomial coefficients and since it is for positive indices it expands only the positive powers.

The general binomial expansion for any index is given by

(x+y)n = nC0xny0 + nC1x(n-1)y1 + nC2x(n-2)y2 +

Answer 2

Binomial Theorem

Binomial Theorems states that for any polynomial the sum of any two terms the nth power can be expressed as a sum of n+1 form.  

Equations gets complicated upon multiplication. Binomial expression helps in reducing the redundancy in calculation using a general formula.  The general form of Binomial theorem is $(a+b)^{n}=\sum_{k=0}^{n} a^{n-k} b^{k}$ where $\left(\begin{array}{c}n \\k\end{array}\right)=\frac{n !}{k !(n-k) !}$.

We will prove Binomial theorem using mathematical induction

Pascals identity can be used, which is in the form of $\left(\begin{array}{c}n \\r-1\end{array}\right)+\left(\begin{array}{l}n \\r\end{array}\right)=\left(\begin{array}{c}n+1 \\r\end{array}\right)$ where $0<r \leq n$$0< r \leq n$.

We need to prove

$(a+b)^{n}=a^{n}+\left(\begin{array}{l}n \\1\end{array}\right) a^{n-1} b+\left(\begin{array}{l}n \\2\end{array}\right) a^{n-2} b^{2}+\cdots+\left(\begin{array}{l}n \\r\end{array}\right) a^{n-r} b^{r}$ .. Eq1

Consider n=1 and n=2 for the above equation for the result to be true.

Replacing n in EQ1 with k, and for all k >2, the equation will be true. This can be written as

$(a+b)^{k}=a^{k}+\left(\begin{array}{c}k \\1\end{array}\right) a^{k-1} b+\left(\begin{array}{c}k \\2\end{array}\right) a^{k-2} b^{2}+\cdots+\left(\begin{array}{c}k \\r\end{array}\right) a^{k-r} b^{r}$

Considering the equation,

$(a+b)^{k+1} = (a+b)(a+b)^{k}$

                $=(a+b)\left(a^{k}+\left(\begin{array}{l}k \\1\end{array}\right) a^{k-1} b\right.+\left(\begin{array}{l}k \\2\end{array}\right) a^{k-2} b^{2}+\cdots+\left(\begin{array}{l}k \\r\end{array}\right) a^{k-r} b^{r}$

                $=a^{k+1}+\left[1+\left(\begin{array}{l}k \\1\end{array}\right)\right] a^{k} b+\left[\left(\begin{array}{l}k \\1\end{array}\right)+\left(\begin{array}{l}k \\2\end{array}\right)\right] a^{k-1} b^{2}+...$

                 $...+\left[\left(\begin{array}{c}k \\r-1\end{array}\right)+\left(\begin{array}{c}k \\r\end{array}\right)\right] a^{k-r+1} b^{r}+\cdots+\left[\left(\begin{array}{c}k \\k-1\end{array}\right)+1\right] a b^{k}+b^{k+1}$

From pascals identity

$(a+b)^{k+1}=a^{k+1}+\left(\begin{array}{c}k+1 \\1\end{array}\right) a^{k} b+\cdots+\left(\begin{array}{c}k+1 \\r\end{array}\right) a^{k-r+1} b^{r}+\cdots+\left(\begin{array}{c}k+1 \\k\end{array}\right) a b^{k}+b^{k+1}$

The result is hence true for k+1.

Thus the result is same for all positive integral index 'n'