Question

State and prove binomial theorem for any positive integer n.

Answer 1

Binomial theorem is proved by principle of mathematical induction. To solve this theorem one should know how to solve combinations and basic identities.
As we all know
$\begin{aligned} & {{\left( x+y \right)}^{2}}={{x}^{2}}+2xy+{{y}^{2}} \\\ & {{\left( x+y \right)}^{3}}={{x}^{3}}+3{{x}^{2}}y+3x{{y}^{2}}+{{y}^{3}} \\\ \end{aligned}$
Here, we can see the power is increasing. For power 2 or 3 we apply the formula but if the power is large, we apply binomial theorem in that scenario.
For example:${{\left( x+y \right)}^{20}},\text{ }{{\left( x+y \right)}^{56}}$ and so on.

Complete step-by-step solution:
Binomial theorem for any positive integer n is :
${{\left( a+b \right)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}{{b}^{0}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+.......+{}^{n}{{C}_{n-1}}{{a}^{1}}{{b}^{n-1}}+{}^{n}{{C}_{n}}{{a}^{0}}{{b}^{n}}$
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\text{ }$ , where n = a non-negative integer and $0\le r\le n$
$\begin{aligned} & {}^{n}{{C}_{n}}=1 \\\ & {}^{n}{{C}_{0}}=1 \\\ \end{aligned}$
Now, let's have a look at the statement of the binomial theorem.
Statement: Binomial theorem states that for any given positive integer n, the expression of the nth power of the sum of any two numbers a and b may take place as the sum of n+1 terms of the particular form.
This statement means when we expand a binomial expression, the sum of terms we get after expansion will be n+1 where n is the power.
Proof of Binomial Theorem:
With the use of mathematical induction, let us prove binomial theorem:
P(n): ${{\left( a+b \right)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}{{b}^{0}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+.......+{}^{n}{{C}_{n-1}}{{a}^{1}}{{b}^{n-1}}+{}^{n}{{C}_{n}}{{a}^{0}}{{b}^{n}}$
For n = 1,
$P\left( 1 \right):\text{ }{{\left( a+b \right)}^{1}}={}^{1}{{C}_{0}}{{a}^{1}}+{}^{1}{{C}_{1}}{{b}^{1}}=a+b,$ which is true.
Just make sure that${{\left( \text{any number} \right)}^{0}}=1$
Let us suppose that P(k) is true for any positive integer k.
Now solve for n = k:
$P\left( k \right):\text{ }{{\left( a+b \right)}^{k}}={}^{k}{{C}_{0}}{{a}^{k}}{{b}^{0}}+{}^{k}{{C}_{1}}{{a}^{k-1}}{{b}^{1}}+{}^{k}{{C}_{2}}{{a}^{k-2}}{{b}^{2}}+.......+{}^{k}{{C}_{k-1}}{{a}^{1}}{{b}^{k-1}}+{}^{k}{{C}_{k}}{{a}^{0}}{{b}^{k}}........(i)$
Now, for k+1:
$P\left( k+1 \right):\text{ }{{\left( a+b \right)}^{k+1}}={}^{k+1}{{C}_{0}}{{a}^{k+1}}{{b}^{0}}+{}^{k+1}{{C}_{1}}{{a}^{k}}{{b}^{1}}+{}^{k+1}{{C}_{2}}{{a}^{k-1}}{{b}^{2}}+.......+{}^{k+1}{{C}_{k+1}}{{a}^{0}}{{b}^{k+1}}$
We can write, ${{\left( a+b \right)}^{k+1}}=\left( a+b \right){{\left( a+b \right)}^{k}}$
So, let’s substitute equation(i) in place of ${{\left( a+b \right)}^{k}}$
Now, we get
$\begin{aligned} & \Rightarrow \left( a+b \right)\left( {}^{k}{{C}_{0}}{{a}^{k}}{{b}^{0}}+{}^{k}{{C}_{1}}{{a}^{k-1}}{{b}^{1}}+{}^{k}{{C}_{2}}{{a}^{k-2}}{{b}^{2}}+.......+{}^{k}{{C}_{k-1}}{{a}^{1}}{{b}^{k-1}}+{}^{k}{{C}_{k}}{{a}^{0}}{{b}^{k}} \right) \\\ & \Rightarrow {}^{k}{{C}_{0}}{{a}^{k+1}}+{}^{k}{{C}_{1}}{{a}^{k}}b+{}^{k}{{C}_{2}}{{a}^{k-1}}{{b}^{2}}.....{}^{k}{{C}_{k-1}}{{a}^{2}}{{b}^{k-1}}+{}^{k}{{C}_{k}}a{{b}^{k}}+{}^{k}{{C}_{0}}{{a}^{k}}b+{}^{k}{{C}_{1}}{{a}^{k-1}}{{b}^{2}}+{}^{k}{{C}_{2}}{{a}^{k-2}}{{b}^{3}}+.....+{}^{k}{{C}_{k-1}}a{{b}^{k}}+{}^{k}{{C}_{k}}{{b}^{k+1}} \\\ \end{aligned}$
Now, group all the like terms:
$\Rightarrow {}^{k}{{C}_{0}}{{a}^{k+1}}+\left( {}^{k}{{C}_{1}}+{}^{k}{{C}_{0}} \right){{a}^{k}}b+\left( {}^{k}{{C}_{2}}+{}^{k}{{C}_{1}} \right){{a}^{k-1}}{{b}^{2}}+.....\left( {}^{k}{{C}_{1}}+{}^{k}{{C}_{k-1}} \right)a{{b}^{k}}+{}^{k}{{C}_{k}}{{b}^{k+1}}$
From above formula of combination,
$\begin{aligned} & \Rightarrow {}^{k}{{C}_{0}}=1 \\\ & \Rightarrow {}^{k+1}{{C}_{0}}=1 \\\ & \Rightarrow {}^{k}{{C}_{r}}+{}^{k}{{C}_{r-1}}={}^{k+1}{{C}_{r}} \\\ & \Rightarrow {}^{k}{{C}_{k}}={}^{k+1}{{C}_{k+1}}=1 \\\ & \Rightarrow {{\left( a+b \right)}^{k+1}}={}^{k+1}{{C}_{0}}{{a}^{k+1}}+{}^{k+1}{{C}_{1}}{{a}^{k}}b+{}^{k+1}{{C}_{2}}{{a}^{k-1}}{{b}^{2}}+.......{}^{k+1}{{C}_{k}}a{{b}^{k}}+{}^{k+1}{{C}_{k+1}}{{b}^{k+1}} \\\ \end{aligned}$
P(k+1) is true whenever P(k) is true.
Therefore, P(n) is true for all positive integral values of n.

Note: Please mark the equation numbers wherever necessary. Do remember all the terms and write them in correct order because in combinations all the terms look similar enough and the expressions should not be repetitive. Do not use any kind of shortcuts as this answer needs to be written step wise.