# Binomial Theorem:

We live in an Era which uses calculation as its base. Since childhood you are studying calculations based on addition, subtraction, multiplication and division. But still, you find it tough to calculate something like this (88)^{5}. It’s simple to say that anyone who knows multiplication can do it but it will take much time. Yes, Of course, you can do it, also you have to save time. To care all these things there was introduced binomial theorem. It provided an easier way to expand (a + b)^{n}, where n is an integer or a rational number. Here, we study binomial theorem for positive integral indices only.

**Binomial Theorem for Positive Integral Indices:**

(a+ b)^{0} = 1 a + b ≠ 0

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

(a+ b)^{2} = a^{2} + 2ab + b^{2}

(a+ b)^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3}

(a+ b)^{4} = (a + b)^{3}(a + b) = a^{4}+ 4a^{3}b + 6a^{2}b^{2} + 4ab^{3} + b^{4}

**Properties of binomial Theorem based on above observation:**

- The number of terms in RHS is 1 more than the terms present in LHS.
- In each successive terms powers of the first quantity ‘a’ go on decreasing by 1 whereas the powers of the second quantity ‘b’ increase by 1.
- In each term of expansion, the sum of powers in each individual term is same as that of original LHS of (a + b)’s power.
- We say the coefficients
^{n}C_{r}occurring in the binomial theorem as binomial coefficients.

**Coefficient of Binomial Expansion:**

Pascal’s Law made it easy to determine the coefficient of binomial expansion. Let’s see:

Suppose,

(a + b)^{5} = 1.a^{4+1} + 5.a^{4}b + 10.a^{3}b^{2} + 10.a^{2}b^{3} + 5.ab^{4} + 1.b^{4+1}

We see here according to Pascal’s triangle that the coefficient of respective terms is same as that obtained in triangle thus you can apply the same everywhere for expansion.

**You can also obtain Pascal’s triangles n**^{th}row by

^{n}C_{0}^{n}C_{1}^{n}C_{2}^{n}C_{3}^{n}C_{……}^{n}C_{n}**Example: For power of 3 i.e. (a + b)**which is same as expansion of^{3}

^{3}C_{0}^{3}C_{1}^{3}C_{2}^{3}C_{3}^{ }1 3 3 1

**(a + b)**a^{3}

e 1.^{3}+**3**.a^{2}b +**3**.ab^{2}+**1.**b^{3}

**Binomial theorem for any positive integer n:**

**(a + b)**^{n}=^{n}C_{0}a^{n}+^{n}C_{1}a^{n–1}b +^{n}C_{2}a^{n–2}b^{2}+ . … . +^{n}C_{n – 1}b^{n–1}+^{n}C_{n}b^{n}

We write it as the sum in short as:**(a + b)**^{n}= \(\sum_{K=0}^{n}\)^{n}C_{k}a^{n-k}b^{k}

**Some Special cases for the expansion of (a + b)**^{n} :

^{n}:

- When we take a = x and b = – y, then

**(x – y)**^{n}=^{n}C_{0}x_{n}–^{n}C_{1}x^{n}^{– 1}y +^{n}C_{2}x^{n – 2}y^{2}+ … + (–1)^{n}^{n}C_{n}y^{n} - When we take a = 1 and b = x, then

**(1 + x)**^{ n}=^{n }C_{0}+^{n}C_{1}x +^{n}C_{2}x^{2}+^{n}C_{3}x^{3}+ … +^{n}C_{n}x^{n}. - When we take a = 1 and b = – x, then

**(1– x)**^{n}=^{n }C_{0}–^{n}C_{1}x +^{n}C_{2}x^{2}–^{n}C_{3}x^{3}+ … + (–1)^{n}^{n}C_{n}x^{n}.

**General and Middle Terms of Binomial Expansion:**

**General term of binomial expansion**: (r +1)^{th }term is said to be general term of expansion:

^{n}C_{r}a^{n–r}.b^{r}**Middle term of binomial expansion :**

- If n is even, then the number of terms in the expansion will be n + 1. Also the middle term will be \(\left( \frac{n}{2} + 1 \right)\)
^{th} - If n is odd, then n +1 is even, so there will be two middle terms in the binomial expansion.

They are: ( \(\frac{n+1}{2}\) )^{th}term and ( \(\frac{n+1}{2} + 1\) )^{th }

**Binomial Theorem Examples:**

to compute*Compute (98)*^{5}.

Solution:

**(98)**. We must express 98 as the sum or difference of two numbers whose powers are easier to handle, and then we will use Binomial Theorem.^{five}

98 = 100 – 2

(98)^{5}= \({(100 – 2)}^{5}\)

=^{5}C_{0} (100)^{4+1} – ^{5}C_{1} (100)^{4} .2 + ^{5}C_{2} (100)^{3} 2^{2} – ^{5}C_{3} (100)^{2} (2)^{3} + ^{5}C_{4} (100) (2)^{4} – ^{5}C_{5} (2)^{5}

= 10000000000 – 5 × 100000000 × 2 + 10 × 1000000 × 4 – 10 ×10000 × 8 + 5 × 100 × 16 – 32

= 10040008000 – 1000800032 = 9039207968

Let’s first break 1.01*Which is larger (1.01)*Solution:^{1000000}or 10,000?

(1.01)^{1000000}

= (1 + 0.01)^{1000000}

= ^{1000000}C_{0} + ^{1000000}C_{1} (0.01) + all the positive terms

= 1 + 1000000 × 0.01 + all the positive terms

= 1 + 10000 + all the positive terms > 10000

Hence, we can say that

(1.01)^{1000000} > 10000

## Be First to Comment