what is the co-efficient of x^98 in (x-1)(x-2)(x-3)..................(x-98)(x-99)(x-100)?

Question

plz help me.............

correction:
(x-1)(x-2)(x-3).................. (x-98)(x-99)(x-100)

Answer 1

n
Π(x - k) = (x - 1)(x - 2)(x - 3) . . . (x - n)
k=1
For the coefficient of the 3rd term you can derive the equation
C3 = (3n^4 + 2n^3 - 3n^2 - 2n)/24
C3 = n(3n^3 + 2n^2 - (3n + 2))/24
C3 = n(n^2(3n + 2) - (3n + 2))/24
C3 = n(n^2 - 1)(3n + 2)/24
C3 = n(n + 1)(n - 1)(3n + 2)/24,
where n is the degree of the product.
C3(100) = 100(101)(99)(302)/24 = 12,582,075

Answer 2

1

Answer 3

What a fascinating question

........(x - 1)(x - 2) = x^2 - 3x + 2
........(x - 2)(x - 3) = x^3 - 6x^2 + 11x - 6
........(x - 3)(x - 4) = x^4 - 10x^3 +35x^2 -50x + 24
........(x - 4)(x - 5) = x^5 - 15x^4 + 85x^3 - 225x^2 ....
........(x - 5)(x - 6) = x^6 - 21x^5 + 175x4 ......
so for
.........(x - 99)(x - 100) we need to look at the coefficient of the 3rd term

this is the sequence 2, 11,35, 85, 175.... and how these numbers were generated in the multiplication
2 = 1 x 2.............................. (x - 2)
11 = 2 + 3 x 3...................(x - 3)
35 = 11 + 6 x 4...................(x - 4)
85 = 35 + 10 x 5.................(x - 5)
175 = 85 + 15 x 6..............(x - 6)
the next one will be 175 + 21 x 7 = 322 .............(x - 7)
the number 21 is from the triangle number sequence n(n+1)/2

For the coefficient of the 3rd term of x - 6 we have
1x2 + 3x3 + 6 x 4 + 10 x 5 + 15 x 6

For the coefficient of the 3rd term of x - 7 we have
1x2 + 3x3 + 6x4 + 10x5 + 15x 6 + 21 x 7
or
1x2x2/2 +2x3x3/2 + 3x4x4/2 + 4x5x2/2 + 5x6x6/2 + 6x7x7/2
or
1x(2^2)/2 + 2x(3^2)/2 + 3x(4^2)/2 +4x(5^2)/2 + 5x(6^2)/2 + 6x(7^2)/2

So for x - 100 you will have the coefficient of x^98 as

1x(2^2)/2 + 2x(3^2)/2 + 3x(4^2)/2.......................99x(100^2)/2

= Sigma from 1 to 99 ( i(i + 1)^2))/2
= Sigma 1 to 99 (i^3 + 2i^2 + i)/2
= [99^2 x 100^2 / 4 + 2x99x100x199/6 + 99x100/2]/2
= 12582075 provided that I haven't made a mistake anywhere!

Answer 4

Think about it logically...

(x-1) = x - 1
(x-1)(x-2) = x^2 - 3x + 2
(x-2)(x-2)(x-3) = x^3 - 6x^2 + 11x - 6
(x-2)(x-2)(x-3)(x-4) = x^4 - 10x^3 + 35x^2 - 50x + 24

now find the pattern!

notice the highest degree remains w/ a coefficient of 1.
notice the 2nd highest degree per line.. how to explain.. the 1st degree has coefficient negative 1 for its 2nd value, the 2nd degree has coefficient negative 3 (1 + 2), the 3rd degree has coefficient negative 6 (1 + 2 + 3), and finally the 4th degree has coefficient negative 10 (1 + 2 + 3 + 4). Continue and you have the coefficient for the 2nd value for the 100th degree (101 * 50 = 5050)

The third value's pattern, although not as easily observable, can be found logically. I'll leave you to that! :)

Answer 5

it's the sum of the double products; which the half difference of the square of the sum with the sum of the squares;

The sum of 1 to 100 is 5050. The sum of the squares from 1 to 100 is

201x101x100/ 6= 6767x50. From there you are all set.

Answer 6

see Viet relations and the 1, 2, 3...100 are the values of roots of polinom.

Answer 7

well the answer is 12582075. too lazy to figure out how right now though