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
iit jee Grade: 12
        

find the coefficient of x^98 in

(x-1)(x-2)(x-3)...(x-99)(x-100)?

7 years ago

Answers : (1)

Ramesh V
70 Points
										


If we expand (x-1)(x-2).......(x-100) we will get

 (x-1)(x-2).......(x-100)=x100 -(1+2+3....100)x99+(Sum of product of all first 100 integers taken two at a time)x98+................

 

Now we have to find "Sum of product of all first 100 integers taken two at a time" which is the required co-efficient.

 

(1+2+3+....100)2=12+22+32+.....+1002+2(Sum of product of all first 100 integers taken two at a time)

 

So  Sum of product of all first 100 integers taken two at a time=((1+2+3+....100)2-(12+22+32+.....+1002))/2

 

Now use 1+2+3+....+n=n(n+1)/2 and 12+22+32+.....+n2=n(n+1)(2n+1)/6

 

Here n=100

 

So we get the required co-efficient=12582075


all the best

--

regards


Ramesh


Now you can win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

7 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete JEE Main/Advanced Course and Test Series
  • OFFERED PRICE: Rs. 15,900
  • View Details
Get extra Rs. 3,180 off
USE CODE: Srinivas20
Get extra Rs. 466 off
USE CODE: Srinivas20

Ask Experts

Have any Question? Ask Experts

Post Question

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