FIND THE SPECIFIC TERM OF BINOMIAL EXPANSION WORKSHEET

Problem 1:

Without simplifying, find:

The 6th term of (2x + 5)15

Solution

Problem 2 :

The 4th term of (x² + 5/x)9

Solution

Problem 3 :

The 10th term of (x - 2/x)17

Solution

Problem 4 :

The 9th term of (2x² - 1/x)21

Solution

Problem 5 :

Find the coefficient of:

x10 in the expansion of (3 + 2x²)10

Solution

Problem 6 :

x3 in the expansion of (2x² - 3/x)6

Solution

Problem 7 :

x12 in the expansion of (2x² - 1/x)12

Solution

Problem 8 :

Find the constant term in:

The expansion of (x + 2/x²)15

Solution

Problem 9 :

The expansion of (x - 3/x²)9

Solution

Answer Key

1)  T6 = 15C5 (2x)10 (5)5

2)  T4 = 9C3 (x²)6 (5/x)3

3)  T10 = 17C9 (x)8 (-2/x)9

4)  T9 = 21C8 (2x²)13 (-1/x)8

5)  So, coefficient of x10 is

10C5 35 25

6)  So, coefficient of x3 is

6C3 23 (-3)3

7)  So, coefficient of x12 is

12C4 28 (-1)4

8)  So, the constant term is

15C5 25

9)  So, the constant term is

9C3 (-3)3

Problem 1:

Find the coefficient of x5 in the expansion of

(x + 3) (2x - 1)6.

Solution

Problem 2 :

Find the coefficient of x5 in the expansion of

(x + 2) (x² + 1)8.

Solution

Problem 3 :

Find the coefficient of x6 in the expansion of

(2 - x) (3x + 1)9

Solution

Problem 4 :

Find the constant term in the expansion of (3x² + 1/x)8.

Solution

Problem 5 :

Find the coefficient of x-6 in the expansion of (2x - 3/x²)12.

Solution

Problem 6 :

Find the coefficient of x5 in the expansion of

(2x + 3) (x - 2)6.

Solution

Problem 7 :

Find the possible values of a if the coefficient of x³ in

(2x + 1/ax²)9 is 288.

Solution

Problem 8 :

Find the term independent of x in the expansion of

(3x - 2/x²)6.

Solution

Answer Key

1)  So, coefficient of x5 is -336.

2)   28

3)  So, coefficient of x6 is 91854.

4)  It does not have one.

5)  So, coefficient of x-6 is 12C6 (2)6 (-3)6.

6) So, coefficient of x5 is 84.

7)  So, value of a is ± 4.

8)  3rd term = 4860.

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