+1 vote
241 views retagged | 241 views
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

by Veteran (102k points)
selected
+1
0

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?

0
yes habib r comming to be in rational form !
0
then $\binom{100}{99}$
0

100 C 1 ??

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

+1
$\binom{100}{99}\equiv \binom{100}{1}$
0

but by the equation,we are getting 100c1  .right??

0

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

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

checked..:-)

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

0
Ya it will be -100
0
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