# Applications of Binomial Expansion

- 2 â‰¤ ,
*n*â‰¥ 1,*n*âˆˆ*N*. - (1 +
*x*)â€“ 1 =^{n}^{n}C_{1}*x*+^{n}C_{2}*x*^{2}+ ... +^{n}C_{n}_{ â€“ 1}*x*^{n}^{ â€“ 1 }+is always divisible by^{n}C_{n}x^{n}*x*also (1 +*x*)â€“ 1 â€“^{n}*nx*=^{n}C_{2}*x*^{2}+ ... +^{n}C_{n}_{ â€“ 1}*x*^{n}^{ â€“ 1 }+is always divisible by^{n}C_{n}x^{n}*x*^{2}.

# Finding remainder using binomial theorem

To find the remainder when

*a*is divided by^{n}*b*we adjust the power of*a*to*a*which is very close to^{m}*b*with difference 1. Also the remainder is always positive. When number of the type 3*k*â€“ 1 is divided by 3, we have =

Hence the remainder is 2.

The following illustration will explain the exact procedure.

**Illustration**

Find the remainder when 5

^{99}is divided by 13.Solution

Here 5

^{2}= 25 which is close to 26 = 13 Ã— 2. Hence*E*= 5

^{99}= 5â‹…5

^{98}= 5â‹…(5

^{2})

^{49}= 5(26 â€“1)

^{49}

â‡’

*E*= 5[^{49}*C*_{0 }26^{49}â€“^{49}*C*_{1 }26^{48}+^{49}*C*_{2 }26^{48}â€“ ... +

^{49}*C*_{48 }26 â€“^{49}*C*_{49}] = 5â‹… 26k â€“ 5Now =

Hence the remainder is 8.