Ch7-Binomial Theorem

In earlier classes, we have learnt how to find the value of \((a+b)^2\) and \((a-b)^3\) which is easy. Using them, we could evaluate the numerical values of numbers like \((98)^2=(100-2)^2,(999)^3=(1000-1)^3\), etc. However, for higher powers like \((98)^5,(101)^6\), etc., the calculations become difficult by using repeated multiplication. This difficulty was overcome by a theorem known as binomial theorem. It gives an easier way to expand \((a+b)^n\), where \(n\) is an integer or a rational number. 

Binomial theorem primarily helps to find the expanded value of the algebraic expression of the form \((a+b)^n\). The calculations get longer and longer as higher exponential values involve too much calculation . But there is some kind of pattern developing.

That pattern is summed up by the Binomial Theorem:

\(
(a+b)^n=\sum_{k=0}^n\left(\begin{array}{l}n \\
k
\end{array}\right) a^{n-k} b^k
\)

Where \(
\left(\begin{array}{l}
n \\
k
\end{array}\right)={ }^n C_k=\frac{(n)(n-1)(n-2) \cdots(n-(k-1))}{k !}=\frac{n !}{k !(n-k) !}
\)

We can also write the theorem as