# HOW TO FIND COEFFICIENT OF X IN BINOMIAL EXPANSION

Example 1 :

Using binomial theorem, indicate which of the following two number is larger: (1.01)100000010000.

Solution :

=  (1.01)1000000 -  10000

=  (1 + 0.01)1000000 -  10000

1000000C1000000C1000000C+ ..........  1000000C1000000 (0.01)1000000 - 10000

=  (1 + 1000000 ⋅ 0.01 + other positive terms) - 10000

=  (1 + 10000 + other positive terms) - 10000

=  1 + other positive terms > 0

0.011000000 >  10000

Example 2 :

Find the coefficient of x15 in  (x2 + (1/x3))10

Solution :

General term  Tr+1  =  nCr x(n-r) ar

x  =  x2, n  =  10, a  =  1/x3

Tr+1  =  nCr x(n-r) ar

=  10Cr x2(10-r) (1/x3)r

=  10Cr x20-2r (x-3r)

=  10Cr x20-5r ------(1)

Now let us find x15 th term

20 - 5r  =  15

20 - 15  =  5r

r  =  5/5  =  1

By applying the value of r in the (1)st equation, we get

=  10C1 x20-5(1)

=  10

Hence the coefficient of  x15 is 10.

Example 3 :

Find the coefficient of x6 and the coefficient of x2 in  (x2 - (1/x3))6

Solution :

General term  Tr+1  =  nCr x(n-r) ar

x  =  x2, n  =  6, a  =  -1/x3

Tr+1  =  nCr x(n-r) ar

=  6Cr x2(6-r) (-1/x3)r

=  6Cr x12-2r (-x-3r)

=  -6Cr x12-5r ------(1)

 Now let us find x6 th term12 - 5r  =  612 - 6  =  5r r  =  6/5 (it is not possible) Now let us find x2  term12 - 5r  =  212 - 2  =  5r r  =  10/5  =  2

x term

=  -6C2 x12-5(2)

=  -15x12-10

=  -15 x2

Coefficient of x2 term is -15.

Example 4 :

Find the coefficient of x4 in the expansion of (1 + x3)50(x2 + 1/x)5.

Solution :

=  (1 + x3)50 (x2 + 1/x)5

=  (1 + x3)55/x5

=  (1 + x3)55/x5

General term  Tr+1  =  nCr x(n-r) ar

x  =  1, n  =  55, a  =  x3

Tr+1  =  nCr x(n-r) ar

=  55Cr (1)(55-r) (x3)r

=  55Cr x3r/x5

=  55Cr x3r-5

3r - 5  =  4

3r  =  9

r  =  3

Coefficient of x4 is 55C3

=  (55 ⋅ 54 ⋅ 53)/(3 ⋅ 2 ⋅ 1)

=  26235

Hence the coefficient of xis 26235.

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