The number of terms in a binomial expansion with an exponent of n is equal to n + 1. To find a particular term in the expansion of (a + b)^{n} we make use of the general term formula.
The general term of the binomial expansion is
T_{r+1} = ^{n}C_{r} a^{nr} b^{r}
Problem 1:
Find the coefficient of x^{5} in the expansion of
(x + 3) (2x  1)^{6}.
Solution :
= (x + 3) (2x  1)^{6}
General term of expansion (a + b)^{n}
T_{r+1} = ^{n}C_{r} a^{nr} b^{r}
Here a = 2x, b = 1, n = 6.
= (x + 3) [^{6}C_{r} (2x)^{6r} (1)^{r}]
= (x + 3) [^{6}C_{r} (2)^{6r} (x)^{6r} (1)^{r}]
= [^{6}C_{r} (2)^{6r} (x)^{6r} (1)^{r} ∙ x] + [^{6}C_{r} (2)^{6r} (x)^{6r} (1)^{r} ∙ 3]
x^{6r} ∙ x = x^{5} x^{7r} = x^{5} 7  r = 5 r = 2 
x^{6 r} = x^{5} 6  r = 5 r = 6  5 r = 1 
= [^{6}C_{2} (2)^{62} (x)^{62} (1)^{2} ∙ x] + [^{6}C_{1} (2)^{61} (x)^{61} (1)^{1} ∙ 3]
= 15(2)^{4} (x)^{5} + 6(2)^{5 }(x)^{5} (1) ∙ 3
= 240x^{5}  576x^{5}
= 336x^{5}
So, coefficient of x^{5} is 336.
Problem 2 :
Find the coefficient of x^{5} in the expansion of
(x + 2) (x² + 1)^{8}.
Solution:
= (x + 2) (x² + 1)^{8}
General term of expansion (a + b)^{n}
T_{r+1} = ^{n}C_{r} a^{nr} b^{r}
Here a = x², b = 1, n = 8.
= (x + 2) [^{8}C_{r} (x²)^{8r} (1)^{r}]
= (x + 2) [^{8}C_{r} (x)^{162r} (1)^{r}]
= [^{8}C_{r} (x)^{162r }(1)^{r} ∙ x] + [^{8}C_{r} (x)^{162r} (1)^{r} ∙ 2] (1)
x^{162r} ∙ x = x^{5} x^{172r} = x^{5} 17  2r = 5 12 = 2r r = 6 
x^{162r} = x^{5} 16  2r = 5 2r = 11 r = 11/2

By applying the value of r in (1), we get
= [^{8}C_{6} (x)^{162(6) }(1)^{6} ∙ x] + [^{8}C_{6} (x)^{162(6)} (1)^{6} ∙ 2]
= [^{8}C_{6} (x)^{5} + [2∙^{8}C_{6} (x)^{4}]
= (8 x 7)/(2 x 1)
= 28
Problem 3 :
Find the coefficient of x^{6} in the expansion of
(2  x) (3x + 1)^{9}
Solution :
= (2  x) (3x + 1)^{9}
General term of expansion (a + b)^{n}
T_{r+1} = ^{n}C_{r} a^{nr} b^{r}
Here a = 3x, b = 1, n = 9.
= (2  x) [^{9}C_{r} (3x)^{9r} (1)^{r}]
= (2  x) [^{9}C_{r} (3)^{9r} (x)^{9r} (1)^{r}]
= [^{9}C_{r} (3)^{9r} (x)^{9r} (1)^{r} ∙ 2] + [^{9}C_{r} (3)^{9r} (x)^{9r} (1)^{r} ∙ (x)]
x^{9r} = x^{6} 9  r = 6 r = 9  6 r = 3 
x^{9 r} ∙ x^{1 }= x^{6} x^{10 r} = x^{6} 10  r = 6 r = 10  6 r = 4 
= [^{9}C_{3} (3)^{93} (x)^{93} (1)^{3} ∙ 2]  [^{9}C_{4} (3)^{94} (x)^{94} (1)^{4} ∙ x]
= 168(3)^{6} (x)^{6}  126 (3)^{5 }(x)^{6}
= 122472x^{6}  30618x^{6}
= 91854x^{6}
So, coefficient of x^{6} is 91854.
Problem 4 :
Find the constant term in the expansion of (3x² + 1/x)^{8}.
Solution :
General term of expansion (a + b)^{n}
T_{r+1} = ^{n}C_{r} a^{nr} b^{r}
Putting a = 3x², b = 1/x, n = 8
T_{r+1} = ^{8}C_{r} (3x²)^{8r} (1/x)^{r}
= ^{8}C_{r} (3)^{8r }(x)^{162r }(1)^{r }(x)^{r}
= ^{8}C_{r} (3)^{8r} (x)^{163r}
Constant term:
16  3r = 0
3r = 16
r = 16/5
It does not have one.
Problem 5 :
Find the coefficient of x^{6} in the expansion of (2x  3/x²)^{12}.
Solution :
= (2x  3/x²)^{12}
General term of expansion (a + b)^{n}
T_{r+1} = ^{n}C_{r} a^{nr} b^{r}
Here a = 2x, b = 3/x², n = 12.
= ^{12}C_{r} (2x)^{12r} (3/x²)^{r}
= ^{12}C_{r} (2)^{12r} (x)^{12r} (3)^{r} (x)^{2r}
= ^{12}C_{r} (2)^{12r} (x)^{123r} (3)^{r}
x^{123r} = x^{6}
12  3r = 6
3r = 18
r = 6
= ^{12}C_{6} (2)^{126} (x)^{1218} (3)^{6}
= ^{12}C_{6} (2)^{6} (3)^{6} (x)^{6}
So, coefficient of x^{6} is ^{12}C_{6} (2)^{6 }(3)^{6}.
Problem 6 :
Find the coefficient of x^{5} in the expansion of
(2x + 3) (x  2)^{6}.
Solution:
= (2x + 3) (x  2)^{6}
General term of expansion (a + b)^{n}
T_{r+1} = ^{n}C_{r} a^{nr} b^{r}
Here a = x, b = 2, n = 6.
= (2x + 3) [^{6}C_{r} (x)^{6r} (2)^{r}]
= [^{6}C_{r} (x)^{6r} (2)^{r} ∙ 2x] + [^{6}C_{r} (x)^{6r} (2)^{r} ∙ 3]
x^{6r} ∙ x = x^{5} x^{7r} = x^{5} 7  r = 5 r = 7  5 r = 2 
x^{6 r} = x^{5} 6  r = 5 r = 6  5 r = 1 
= [^{6}C_{2} (x)^{62} (2)^{2} ∙ 2x] + [^{6}C_{1} (x)^{61} (2)^{1} ∙ 3]
= 120x^{5}  36x^{5}
= 84x^{5}
So, coefficient of x^{5} is 84.
Problem 7 :
Find the possible values of a if the coefficient of x³ in
(2x + 1/ax²)^{9} is 288.
Solution :
= (2x + 1/ax²)^{9}
General term of expansion (a + b)^{n}
T_{r+1} = ^{n}C_{r} a^{nr} b^{r}
Here a = 2x, b = 1/ax², n = 9.
= ^{9}C_{r} (2x)^{9r} (1/ax²)^{r}
= ^{9}C_{r} (2x)^{9r} (1/a)^{r} (x)^{2r}
= ^{9}C_{r} (2)^{9r} (x)^{93r} (1/a)^{r }
x^{93r} = x^{3}
9  3r = 3
3r = 6
r = 2
= ^{9}C_{2} (2)^{92} (x)^{96} (1/a)^{2}
= ^{9}C_{2} (2)^{7} (x)^{3} (1/a)^{2}
288 = 4608 x³ (1/a)²
(1/a)² = 4608/288
(1/a)² = 16
a = ± 4
So, value of a is ± 4.
Problem 8 :
Find the term independent of x in the expansion of
(3x  2/x²)^{6}.
Solution :
= (3x  2/x²)^{6}
General term of expansion (a + b)^{n}
T_{r+1} = ^{n}C_{r} a^{nr} b^{r}
Here a = 3x, b = 2/x², n = 6.
= ^{6}C_{r} (3x)^{6r} (2/x²)^{r}
= ^{6}C_{r} (3)^{6r} (x)^{6r} (2)^{r} (x)^{2r}
T_{r+1} = ^{6}C_{r} (3)^{6r} (x)^{63r} (2)^{r}
To find the term independent of x. So, power of x is 0.
x^{63r} = 0
6  3r = 0
3r = 6
r = 2
T_{2+1} = ^{6}C_{2} (3)^{62} (x)^{66} (2)^{2}
T_{3} = ^{6}C_{2} (3)^{4} (x)^{0} (2)^{2}
= 4860
Hence, the term which is independent of x is
3^{rd} term = 4860.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM