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Binomial Theorem

This page provides an introduction to Statistics.

Overview​

The Binomial Theorem is a mathematical formula that describes the expansion of powers of a binomial (a polynomial with two terms) into a sum of terms involving various powers of the individual terms.

Binomial Coefficient​

Formula for binomial coefficient is as below:

nCr=n!r!(n - r)!^\text{n}\text{C}_\text{r} = \frac{\text{n}!}{\text{r}! \text{(n - r)}!}

Here are some of the characteristics of binomial coefficient.

  • nC0  =  nCn=1^\text{n}\text{C}_0 \ \ = \ \ ^\text{n}\text{C}_\text{n} = \text{1}

  • nC1  =  nC(n - 1)=n^\text{n}\text{C}_1 \ \ = \ \ ^\text{n}\text{C}_\text{(n - 1)} = \text{n}

  • nCr  =  nC(n - r)^\text{n}\text{C}_\text{r} \ \ = \ \ ^\text{n}\text{C}_\text{(n - r)}

  • Sum of all binomial coefficients will be:

∑r=0n  nCr=2n\sum_{\text{r} = 0}^\text{n} \ \ ^\text{n}\text{C}_\text{r} = 2^{\text{n}}
  • Sum of all binomial coefficients whose lower suffix is even will be:
nC0  +  nC2  +  nC4  +  ...  =  2n - 1^\text{n}\text{C}_0 \ \ + \ \ ^\text{n}\text{C}_2 \ \ + \ \ ^\text{n}\text{C}_4 \ \ + \ \ ...\ \ = \ \ 2^\text{{n - 1}}
  • Sum of all binomial coefficients whose lower suffix is odd will be:
nC1  +  nC3  +  nC5  +  ...  =  2n - 1^\text{n}\text{C}_1 \ \ + \ \ ^\text{n}\text{C}_3 \ \ + \ \ ^\text{n}\text{C}_5 \ \ + \ \ ... \ \ = \ \ 2^\text{{n - 1}}

  • nCr  +  nCr + 1  =  n + 1Cr + 1^\text{n}\text{C}_\text{r} \ \ + \ \ ^\text{n}\text{C}_\text{{r + 1}} \ \ = \ \ ^\text{{n + 1}}\text{C}_\text{{r + 1}}

  • nCr  +  nCr - 1  =  n + 1Cr^\text{n}\text{C}_\text{r} \ \ + \ \ ^\text{n}\text{C}_\text{{r - 1}} \ \ = \ \ ^\text{{n + 1}}\text{C}_\text{r}

  • pCq  +  p+1Cq  +  p+2Cq  +  ...  +  p+nCq  =  p+n+1Cq+1^\text{p}\text{C}_\text{q} \ \ + \ \ ^\text{{p+1}}\text{C}_\text{q} \ \ + \ \ ^\text{{p+2}}\text{C}_\text{q} \ \ + \ \ ... \ \ + \ \ ^\text{{p+n}}\text{C}_\text{q} \ \ = \ \ ^\text{{p+n+1}}\text{C}_\text{{q+1}}

  • r  ∗  nCr  =  n  ∗  n - 1Cr - 1\text{r} \ \ * \ \ ^\text{n}\text{C}_\text{r} \ \ = \ \ \text{n} \ \ * \ \ ^\text{{n - 1}}\text{C}_\text{{r - 1}}

Max Value of Binomial Coefficient​

Max value of nCr^\text{n}\text{C}_\text{r} is when

  • r=n2\text{r} = \large\frac{\text{n}}{2}, and n\text{n} is even.

  • r=n+12\text{r} = \large\frac{\text{n} + 1}{2} or r=n - 12\text{r} = \large\frac{\text{n - 1}}{2}, and n\text{n} is odd.

You can relate it with pascal's triangle.

  • 5C0=1^5\text{C}_0 = 1
  • 5C1=5^5\text{C}_1 = 5
  • 5C2=10^5\text{C}_2 = 10, maximum value
  • 5C3=10^5\text{C}_3 = 10, maximum value
  • 5C4=5^5\text{C}_4 = 5
  • 5C5=1^5\text{C}_5 = 1

Binomial Expansion​

(a+b)n=(nC0∗an∗b0)  +  (nC1∗an−1∗b1)  +  (nC2∗an−2∗b2)  +  ....  +  (nCn∗a0∗bn)(\text{a} + \text{b})^\text{n} = (^\text{n}\text{C}_0 * \text{a}^\text{n} * \text{b}^0) \ \ + \ \ (^\text{n}\text{C}_1 * \text{a}^{\text{n} - 1} * \text{b}^1) \ \ + \ \ (^\text{n}\text{C}_2 * \text{a}^{\text{n} - 2} * \text{b}^2) \ \ + \ \ .... \ \ + \ \ (^\text{n}\text{C}_\text{n} * \text{a}^0 * \text{b}^\text{n})

It can also be written as below:

(a+b)n=∑r=0n  nCr  ∗  an−r  ∗  br(\text{a} + \text{b})^\text{n} = \sum_{\text{r} = 0}^\text{n} \ \ ^\text{n}\text{C}_\text{r} \ \ * \ \ \text{a}^{\text{n} - \text{r}} \ \ * \ \ \text{b}^\text{r}

Important Points​

  • Formula for Tr + 1\text{T}_\text{{r + 1}} of Binomial Expansion
Tr + 1=  nCr  ∗  an - r  ∗  br\text{T}_\text{{r + 1}} = \ \ ^\text{n}\text{C}_\text{r} \ \ * \ \ \text{a}^\text{{n - r}} \ \ * \ \ \text{b}^\text{r}

  • There are total (n + 1)\text{(n + 1)} terms in binomial expansion of (a + b)n\text{(a + b)}^\text{n}

  • Middle term of (a + b)n\text{(a + b)}^\text{n} is:

    • (n+22)th\large(\frac{\text{n} + 2}{2})^{\text{th}}, if n\text{n} is even.

    • (n+12)th\large(\frac{\text{n} + 1}{2})^{\text{th}} and (n+32)th\large(\frac{\text{n} + 3}{2})^{\text{th}}, if n\text{n} is odd.

Additional Binomial Expansion​

Let's say we have equation 11 as below:

(a+b)n=(nC0∗an∗b0)  +  (nC1∗an−1∗b1)  +  (nC2∗an−2∗b2)  +  ....  +  (nCn∗a0∗bn)(\text{a} + \text{b})^\text{n} = (^\text{n}\text{C}_0 * \text{a}^\text{n} * \text{b}^0) \ \ + \ \ (^\text{n}\text{C}_1 * \text{a}^{\text{n} - 1} * \text{b}^1) \ \ + \ \ (^\text{n}\text{C}_2 * \text{a}^{\text{n} - 2} * \text{b}^2) \ \ + \ \ .... \ \ + \ \ (^\text{n}\text{C}_\text{n} * \text{a}^0 * \text{b}^\text{n})

we have another equation 22 as below:

(a−b)n=(nC0∗an∗b0)  −  (nC1∗an−1∗b1)  +  (nC2∗an−2∗b2)  −  ....  +  (nCn∗a0∗bn)(\text{a} - \text{b})^\text{n}= (^\text{n}\text{C}_0 * \text{a}^\text{n} * \text{b}^0) \ \ - \ \ (^\text{n}\text{C}_1 * \text{a}^{\text{n} - 1} * \text{b}^1) \ \ + \ \ (^\text{n}\text{C}_2 * \text{a}^{\text{n} - 2} * \text{b}^2) \ \ - \ \ .... \ \ + \ \ (^\text{n}\text{C}_\text{n} * \text{a}^0 * \text{b}^\text{n})

Adding equation 11 and 22 we get:

(a+b)n+(a−b)n=2∗[(nC0∗an∗b0)  +  (nC2∗an−2∗b2)  +  (nC4∗an−4∗b4)](\text{a} + \text{b})^\text{n} + (\text{a} - \text{b})^\text{n} = 2 * [(^\text{n}\text{C}_0 * \text{a}^\text{n} * \text{b}^0) \ \ + \ \ (^\text{n}\text{C}_2 * \text{a}^{\text{n} - 2} * \text{b}^2) \ \ + \ \ (^\text{n}\text{C}_4 * \text{a}^{\text{n} - 4} * \text{b}^4)]

Example 1​

Finding remainder using binomial expansion.

32008\frac{3^{200}}{8}

Solution​

First find which power of 33 that is close to 88, it will be 22 because 32=93^2 = 9, then re-write the equation as below:

(32)1008\frac{(3^{2})^{100}}{8}

It can also be written as below:

(8+1)1008\frac{(8 + 1)^{100}}{8}

Using binomial expansion we get:

[(100C0∗8100)+(100C1∗899)+(100C2∗898)+....+100C100∗80]  ÷  8[(^{100}\text{C}_0 * 8^{100}) + (^{100}\text{C}_1 * 8^{99}) + (^{100}\text{C}_2 * 8^{98}) + .... + ^{100}\text{C}_{100} * 8^{0}] \ \ \div \ \ 8

Simplifying it futher we get:

[(100C0∗899)+(100C1∗898)+(100C2∗897)+....+100C99∗80]  +  100C100 ∗ 808[(^{100}\text{C}_0 * 8^{99}) + (^{100}\text{C}_1 * 8^{98}) + (^{100}\text{C}_2 * 8^{97}) + .... + ^{100}\text{C}_{99} * 8^{0}] \ \ + \ \ \frac{^{100}C_{100} \ * \ 8^{0}}{8} [(100C0∗899)+(100C1∗898)+(100C2∗897)+....+100C99∗80]  +  18[(^{100}\text{C}_0 * 8^{99}) + (^{100}\text{C}_1 * 8^{98}) + (^{100}\text{C}_2 * 8^{97}) + .... + ^{100}\text{C}_{99} * 8^{0}] \ \ + \ \ \frac{1}{8}

So remainder is 11 as our term before 18\large\frac{1}{8} already got divided by 88.