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Mathematical Induction & Binomial Theorem¶

An algebraic expression consisting of two terms such as $a+x$, $ax+b$ etc is called binomial or a binomial expression.

Binomial Formula¶

The following expressions are valid for any $n\in Z^n$. (n is called index)

\[ (a+b)^n = a^n +\frac{n}{1!} a^{n-1} b^1+ \frac{n(n-1)}{2!} a^{n-2} b^2 + \frac{n(n-1)(n-2)}{3!} a^{n-3} b^3 +...+ b^n \]

\[ (a+b)^n = ^nC_0 a^n + ^nC_1 a^{n-1} b^1 + ^nC_2 a^{n-2} b^2 +...+ ^nC_r a^{n-r} b^r +...+^nC_n a^{n-n} b^n \]

\[ (a+b)^n = \displaystyle\sum_{r=0}^n \binom{n}{r} a^{n-r} b^r \]

The number of terms in the expansion is one greater than its index.
The sum of expansion of $a$ and $b$ in each term of the expansion is equal to its index.
The exponent of $a$ decreases from index to zero.
The exponent of $b$ increases form zero to index.
The coefficients of the term equidistant from beginning and end of the expansion are equal as $^nC_r=^nC_{n-r}$
The $T_{r+1}$ is the $r^{\text{th}}$ teem in the expansion of $(a+b)^n$ is given by $$ T_{r+1}= \binom{n}{r} a^{n-r} b^r $$
Middle term(s) in the expansion of $(a+b)^n$
- If $n$ is even, then there is only one middle term $\left( \frac{n+2}{2} \right)$th term.
- If n is odd, then there are two middle terms $\left(\frac{n+1}{2}\right)$ and $\left(\frac{n+3}{2}\right)$
In the expansion $$ (a+b)^{n = a}n +\frac{n}{1!} a^{n-1} b^1+ \frac{n(n-1)}{2!} a^{n-2} b^2 + \frac{n(n-1)(n-2)}{3!} a^{n-3} b^3 +...+ b^n $$ $^nC_0, ^nC_1, ^nC_2, ..., ^nC_n$ are called binomial coefficients.

In the expansion of $(1+x)^n$ where $n\in N$

Sum of binomial coefficients is $2^n$ i.e. $^nC_0, ^nC_1, ^nC_2, ..., ^nC_n=2^n$
Sum of binomial coefficients of odd terms = Sum of binomial coefficients of even terms

When n is negative integer or a fraction, and $|x<1|$ then $$ (1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3+...+ \frac{n(n-1)(n-2)...(n-r+1)}{r!}x^r+...$$
The general term in the expansion is $$ T_{r+1}=\frac{n(n-1)(n-2)...(n-r+1)}{r!}x^r $$
The number of terms in the expansion is always infinite.