# find the coffecient of x^198

1 vote
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retagged
0

1 vote

Use Binomial Theorem simply ,

So rth term in the given binomial expansion will be :

(-1)r  nCr (x2)n-r  (1/x)r  ; here n = 100

==>   (-1)r . nCx200 - 3r    =   x198

But equating 200 - 3r and 198 we get r = 2 / 3 which is not an integral value..

But we require that r be an integer as it is used in binomial coefficient nCr..

Hence there is no term containing x198..

So coefficient of x198 in the given binomial expansion = 0

selected
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of they had asked for x197 then coefficent is there right?because it is of the form 22 -3r.

so what would have been the answer?

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yes habib r comming to be in rational form !
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then $\binom{100}{99}$
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100 C 1 ??

i know 100c99 and 100c1 are same things but by formula we get 100c1..right?

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$\binom{100}{99}\equiv \binom{100}{1}$
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but by the equation,we are getting 100c1  .right??

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or is it - (100c1) because of (-1) r. ..??

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Hence since no integral r value exists hence 0 will be the answer..Ok let me edit it now..:)
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@Akriti sood , plz check now..
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checked..:-)

can you confirmthe coefficent of x197 ??is it (- 1)(100c1)?

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Ya it will be -100
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bro!!!  ...can you solve this equation to find formula for the coefficient of x^k

this ques is from kenneth h rosen ex-5.4 question 11

my answer is not matching with the book solution can you explain it

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