IIT JEE , AIEEE Forum - Mathematics , Algebra - For IIT-JEE (Beginers)

shashankashekhar.dutta

Find the co-efficient of x⁹⁸ in (x-1)(x-2)(x-3)....(x-99)(x-100)

aansh_c

hy dis 1 is a simple ques

c basically understnd d logic behind it

(x-a1)(x-a2)

x^2 - x*(a1+a2)+a1*a2

similarly

(x-a1)(x-a2)(x-a3)

x^3-x^2(a1+a2+a3)+x(a1*a2+a2*a3+a3*a1)-a1*a2*a3

nw 4 ur question

d coefficent of x^98 will b

(1.2+2.3+3.4+...........+99.100)

4 dis u'll hv 2 calculate 1.2+2.3+..........+(n-1).n

nth term will be n^2-n

we hv 2 calculate sigma nth term

v hv

[n*(n+1)*(2*n+1)]/6 + [n*(n+1)]/2

den put n=100

[100*101*201]/6 + [100*101]/2

338350+5050

343400

iitkgp_bipin

If the expression is in the form :

(x-x₁)(x-x₂).......(x-x_n) = xⁿ + a₁x^n-1 + a₂x^n-2 + ........ a_n-1x + a_n

Then the coefficient of x^n-r = (-1)^r(product of the roots taken 'r' at a time).

Here (x-1)(x-2).......(x-99)(x-100)

Coefficient of x⁹⁸ = (-1)²(product of the roots taken 2 at a time)

= (1.2 + 2.3 + 3.4 + ...... + 99.100)

= _[r=1]

^[r=99] r(r+1)

= _[r=1]

^[r=99] r² + r

= N(N+1)(2N+1)/6 + N(N+1)/2 where N = 99

nadeemoidu

The product of roots taken 2 at a time in the above answer is wrong.

the correct way is ( 1.2 + 1.3 + 1.4 .... + 2.3 + 2.4 + 2.5 +.......99.100)

To find this ,
consider ( 1+ 2+ 3+ ..... 100)( 1+ 2+ 3+ ..... 100) = 5050*5050=25502500

here the squares of the nos. have been added as extra and all the other terms have repeated exactly twice.

so subtract the sum of squares and divide by 2.

answer = [ 25502500 - ( 100) ( 101) ( 201) / 6 ]/2= 12582075

Ask & Discuss Questions with Community & Experts