SOLUTION: Help is urgently needed.
Find the remainder when x^100 + 2x^99 +3x^98 +... 100x is divided by {{{ x-1 }}}
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Question 995916: Help is urgently needed.
Find the remainder when x^100 + 2x^99 +3x^98 +... 100x is divided by
Found 2 solutions by Boreal, ikleyn:
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
synthetic division
1/1+2+3+4+5...
=1+1
===3 +3
======
=====6 +4
The first four terms are
x^99+3x^98+6x^97+10x^96+...
The coefficient of each term is the sum of all the preceding terms plus the coefficient of the next term.
The pattern will be 15,21,28,36.45.55.....
The coefficient of the 100th term is the sum of the preceding 99 terms plus the coefficient of the 100th term (100), so the quotient will be dividing x-1 into 4950x+100x or 5050 x.
That will go in 5050 times, and we subtract 5050x-5050, so that the remainder is 5050.
Answer by ikleyn(52814) (Show Source): You can put this solution on YOUR website!
.
The remainder theorem says (see the lesson Divisibility of polynomial f(x) by binomial x-a in this site) that the remainder of division the polynomial by
the binomial is equal to the value of the polynomial.
By applying this theorem to the given polynomial f(x) = , it means that the remainder from division the polynomial by is equal
to the value of the polynomial.
In turn, this value is
f(1) = = 1 + 2 + 3 + . . . + 100.
It is the sum of first 100 natural numbers = sum of 100 terms of the arithmetic progression with the first term 1 and the common difference 1.
This sum is equal to = 50*101 = 5050.
So, the required remainder is 5050.
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