Loading....
Coupon Accepted Successfully!

 

Applications of Binomial Expansion

  • 2 ≤ 67125.png, n ≥ 1, n N.
  • (1 + x)n – 1 = nC1x + nC2 x2 + ... + nCn – 1xn – 1 + nCnxn is always divisible by x also (1 + x)n – 1 – nx = nC2 x2 + ... + nCn – 1xn – 1 + nCnxn is always divisible by x2.

Finding remainder using binomial theorem

To find the remainder when an is divided by b we adjust the power of a to am which is very close to b with difference 1. Also the remainder is always positive. When number of the type 3k – 1 is divided by 3, we have
 
67119.png = 67112.png
 
Hence the remainder is 2.
 
The following illustration will explain the exact procedure.
 
Illustration
Find the remainder when 599 is divided by 13.
Solution
Here 52 = 25 which is close to 26 = 13 × 2. Hence
E = 599 = 5⋅598 = 5⋅(52)49 = 5(26 –1)49
 
⇒ E = 5[49C0 2649 – 49C1 2648 + 49C2 2648
– ... + 49C48 26 – 49C49] = 5⋅ 26k – 5
 
Now 67106.png = 67100.png
 
Hence the remainder is 8.
 




Test Your Skills Now!
Take a Quiz now
Reviewer Name