MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Grade: 12
        
Please solve the problem from binomial theorem.
1)   Find the coefficient of x^65 in the expansion of x(x+1)^2(x+2)^3(x+3)^4.........(x+10)^11
2).  Find the coefficient of x^(n^2+n-14)/2 in the expansion of (x-1)(x^2-2)(x^3 -3).........(x^n-n)
9 months ago

Answers : (1)

Aditya Gupta
1670 Points
							
Roots of x(x+1)^2...(x+10)^11 are
0, -1, -1, -2, -2, -2,..., -10, -10,...,-10
Also, the degree of the above polynomial is 1+2+...+11= 66. So coefficient of x^65 is the following sum:
∑r(r+1), r ranging from 0 to 10. This is because of viettas formulas taking the roots one at a time.
The above sum is easy to find coz ∑r^2 and ∑r already have standard formulas
9 months ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

  • 731 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 101 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details