RD Chapter 18 Binomial Theorem Ex 18.2 Solutions
Question - 1 : - Find the 11th term from the beginning and the 11th termfrom the end in the expansion of (2x – 1/x2)25.
Answer - 1 : -
Given:
(2x – 1/x2)25
The given expressioncontains 26 terms.
So, the 11th termfrom the end is the (26 − 11 + 1) th term from thebeginning.
In other words, the 11th termfrom the end is the 16th term from the beginning.
Then,
T16 =T15+1 = 25C15 (2x)25-15 (-1/x2)15
= 25C15 (210)(x)10 (-1/x30)
= – 25C15 (210 /x20)
Now we shall find the11th term from the beginning.
T11 =T10+1 = 25C10 (2x)25-10 (-1/x2)10
= 25C10 (215)(x)15 (1/x20)
= 25C10 (215 /x5)
Question - 2 : - Find the 7th term in theexpansion of (3x2 – 1/x3)10.
Answer - 2 : -
Given:
(3x2 –1/x3)10
Let us consider the 7th termas T7
So,
T7 = T6+1
= 10C6 (3x2)10-6 (-1/x3)6
= 10C6 (3)4 (x)8 (1/x18)
= [10×9×8×7×81] /[4×3×2×x10]
= 17010 / x10
∴ The 7th termof the expression (3x2 – 1/x3)10 is17010 / x10.
Question - 3 : - Find the 5th term in the expansion of (3x – 1/x2)10.
Answer - 3 : -
Given:
(3x – 1/x2)10
The 5th termfrom the end is the (11 – 5 + 1)th, is., 7th term from thebeginning.
So,
T7 = T6+1
= 10C6 (3x)10-6 (-1/x2)6
= 10C6 (3)4 (x)4 (1/x12)
= [10×9×8×7×81] /[4×3×2×x8]
= 17010 / x8
∴ The 5th termof the expression (3x – 1/x2)10 is 17010 / x8.
Question - 4 : - Find the 8th term in the expansion of (x3/2 y1/2 –x1/2 y3/2)10.
Answer - 4 : -
Given:
(x3/2 y1/2 –x1/2 y3/2)10
Let us consider the 8th termas T8
So,
T8 = T7+1
= 10C7 (x3/2 y1/2)10-7 (-x1/2 y3/2)7
= -[10×9×8]/[3×2] x9/2 y3/2 (x7/2 y21/2)
= -120 x8y12
∴ The 8th termof the expression (x3/2 y1/2 – x1/2 y3/2)10 is-120 x8y12.
Question - 5 : - Find the 7th term in the expansion of (4x/5 + 5/2x) 8.
Answer - 5 : -
Given:
(4x/5 + 5/2x) 8
Let us consider the 7th termas T7
So,
T7 = T6+1
∴ The 7th termof the expression (4x/5 + 5/2x) 8 is 4375/x4.
Question - 6 : - Find the 4th term from the beginning and 4th termfrom the end in the expansion of (x + 2/x) 9.
Answer - 6 : -
Given:
(x + 2/x) 9
Let Tr+1 bethe 4th term from the end.
Then, Tr+1 is(10 − 4 + 1)th, i.e., 7th, term from the beginning.
Question - 7 : - Find the 4th term from the end in the expansion of (4x/5– 5/2x) 9.
Answer - 7 : -
Given:
(4x/5 – 5/2x) 9
Let Tr+1 bethe 4th term from the end of the given expression.
Then, Tr+1 is(10 − 4 + 1)th term, i.e., 7th term, from the beginning.
T7 = T6+1
∴ The 4th termfrom the end is 10500/x3.
Question - 8 : - Find the 7th term from the end in the expansion of (2x2 –3/2x) 8.
Answer - 8 : -
Given:
(2x2 –3/2x) 8
Let Tr+1 bethe 4th term from the end of the given expression.
Then, Tr+1 is(9 − 7 + 1)th term, i.e., 3rd term, from the beginning.
T3 = T2+1
∴ The 7th termfrom the end is 4032 x10.
Question - 9 : - Find the coefficient of:
(i) x10 in the expansion of (2x2 –1/x)20
(ii) x7 in the expansion of (x – 1/x2)40
(iii) x-15 in the expansion of (3x2 –a/3x3)10
(iv) x9 in the expansion of (x2 – 1/3x)9
(v) xm in the expansion of (x + 1/x)n
(vi) x in the expansion of (1 – 2x3 + 3x5) (1+ 1/x)8
(vii) a5b7 in the expansion of (a – 2b)12
(viii) x in the expansion of (1 – 3x + 7x2) (1 – x)16
Answer - 9 : -
(i) x10 inthe expansion of (2x2 – 1/x)20
Given:
(2x2 –1/x)20
If x10 occursin the (r + 1)th term in the given expression.
Then, we have:
Tr+1 = nCr xn-r ar
(ii) x7 inthe expansion of (x – 1/x2)40
Given:
(x – 1/x2)40
If x7 occursat the (r + 1) th term in the given expression.
Then, we have:
Tr+1 = nCr xn-r ar
40 − 3r =7
3r = 40 – 7
3r = 33
r = 33/3
= 11
(iii) x-15 inthe expansion of (3x2 – a/3x3)10
Given:
(3x2 –a/3x3)10
If x−15 occursat the (r + 1)th term in the given expression.
Then, we have:
Tr+1 = nCr xn-r ar
(iv) x9 inthe expansion of (x2 – 1/3x)9
Given:
(x2 –1/3x)9
If x9 occursat the (r + 1)th term in the above expression.
Then, we have:
Tr+1 = nCr xn-r ar
For this term to contain x9, we must have:
18 − 3r = 9
3r = 18 – 9
3r = 9
r = 9/3
= 3
(v) xm inthe expansion of (x + 1/x)n
Given:
(x + 1/x)n
If xm occursat the (r + 1)th term in the given expression.
Then, we have:
Tr+1 = nCr xn-r ar
(vi) x in the expansion of(1 – 2x3 + 3x5) (1 + 1/x)8
Given:
(1 – 2x3 +3x5) (1 + 1/x)8
If x occursat the (r + 1)th term in the given expression.
Then, we have:
(1 – 2x3 +3x5) (1 + 1/x)8 = (1– 2x3 + 3x5) (8C0 + 8C1 (1/x)+ 8C2 (1/x)2 + 8C3 (1/x)3 + 8C4 (1/x)4 + 8C5 (1/x)5 + 8C6 (1/x)6 + 8C7 (1/x)7 + 8C8 (1/x)8)
So, ‘x’ occurs in theabove expression at -2x3.8C2 (1/x2)+ 3x5.8C4 (1/x4)
∴ Coefficient of x = -2(8!/(2!6!)) + 3 (8!/(4! 4!))
= -56 + 210
= 154
(vii) a5b7 inthe expansion of (a – 2b)12
Given:
(a – 2b)12
If a5b7 occursat the (r + 1)th term in the given expression.
Then, we have:
Tr+1 = nCr xn-r ar
(viii) x in the expansion of(1 – 3x + 7x2) (1 – x)16
Given:
(1 – 3x + 7x2)(1 – x)16
If x occursat the (r + 1)th term in the given expression.
Then, we have:
(1 – 3x + 7x2)(1 – x)16 = (1 – 3x + 7x2) (16C0 + 16C1 (-x)+ 16C2 (-x)2 + 16C3 (-x)3 + 16C4 (-x)4 + 16C5 (-x)5 + 16C6 (-x)6 + 16C7 (-x)7 + 16C8 (-x)8 + 16C9 (-x)9 + 16C10 (-x)10 + 16C11 (-x)11 + 16C12 (-x)12 + 16C13 (-x)13 + 16C14 (-x)14 + 16C15 (-x)15 + 16C16 (-x)16)
So, ‘x’ occurs in theabove expression at 16C1 (-x) – 3x16C0
∴ Coefficient of x =-(16!/(1! 15!)) – 3(16!/(0! 16!))
= -16 – 3
= -19
Question - 10 : - Which term in the expansion of 
contains x and y to one and the same power.
Answer - 10 : -
Let us consider Tr+1 thterm in the given expansion contains x and y to one and the same power.
Then we have,
Tr+1 = nCr xn-r ar
