IIT JEE , AIEEE Forum - Mathematics , Algebra - what is the co-efficient of x^98 in (x-1)(x-2)(x-3)..................(x-98)(x-99)(x-100)?"

nano0101

PLz help me with this problem........ give me a detailed solving plzzzzzzz......

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

chirag

Answer.:-343400

for x^98, we take the double product since it is the third last term..

i.e.

1*2+2*3+3*4+4*5+............................99*100

applying summation, we get the answer....

nitin62225

@ above it's undercounting!

nano0101

no the answer shud be 12582075 !!!

My problem is that i did'nt get the funda !!! how exactly can we solve this prob???

nitin62225

ans= h(101-h)+101 [ _{[ n=k]}

^{[101-k ]} (n)]

^{h=1,2,3.....50} ^{k=1,2,3,....49}

soln:

(x²-101x+100)(x²-101x+198)(x-3)(x-4).......(x-98)

(x²-101x+97*3)(x²-101x+96*4)(x-5)(x-6)......(x-95)

proceeding like wise, the ans above mentioned is obtained.

plz rate.......(if incorrect then plz point)

nano0101

@ above ......... thats over my head to understand exactly !!!!

I want just the method ........ to understand not to just blindly arrive at the answer !!

anyways thanks ...........

LastMinuteGenius

for this problem, what you need to do is take the x's from 98 of those product terms (ie. a product term is [x-1] or [x-2],etc) and take the constants from the other 2 terms, so, it is what chirag says:
1*2 + 2*3 + ... + 98*99 + 99*100
general term = r(r+1)
so this is r(r+1)[(r+2) - (r-1)]/3
=> (1/3)*[r(r+1)(r+2) - (r-1)r(r+1)]
summing r from 1 to 99, we get
=> (1/3)*[1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 so on]
as all get cancelled, we get
=> (1/3)*[99.100.101]

so, answer is 33*100*101 => 333,300

RATE if im correct...

nano0101

nope!! as i told earlier the answer shud be 12582075 !!!

nitin62225

ans= h(101-h)+101 [ _{[ n=k]}

^{[101-k ]} (n)]

^{h=1,2,3.....50} ^{k=1,2,3,....49}

soln:

(x²-101x+100)(x²-101x+198)(x-3)(x-4).......(x-98)

(x²-101x+97*3)(x²-101x+96*4)(x-5)(x-6)......(x-95)

proceeding like wise, the ans above mentioned is obtained.

plz rate.......(if incorrect then plz point)

nano0101

nope hehehe ......... i guess the method has to be a lil' simple cos this problem comes under "not so difficult" kind of questions........

nitin62225

@ i hve rechecked my soln and cannot find nething wrong! so if u can please show me the fault itll b nice of u!

and if u hve a better(easier) method, do tell me i m anxious 2 know!

KAB

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

nitin62225

thnx!

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