RD Chapter 18 Binomial Theorem Ex 18.2 Solutions
Question - 11 : - Does the expansion of (2x2 – 1/x) contain any term involving x9?
Answer - 11 : -
Given:
(2x2 –1/x)
If x9 occursat the (r + 1)th term in the given expression.
Then, we have:
Tr+1 = nCr xn-r ar
For this term tocontain x9, we must have
40 – 3r = 9
3r = 40 – 9
3r = 31
r = 31/3
It is not possible,since r is not an integer.
Hence, there is noterm with x9 in the given expansion.
Question - 12 : - Show that the expansion of (x2 + 1/x)12 does not contain any term involving x-1.
Answer - 12 : -
Given:
(x2 +1/x)12
If x-1 occursat the (r + 1)th term in the given expression.
Then, we have:
Tr+1 = nCr xn-r ar
For this term tocontain x-1, we must have
24 – 3r = -1
3r = 24 + 1
3r = 25
r = 25/3
It is not possible,since r is not an integer.
Hence, there is noterm with x-1 in the given expansion.
Question - 13 : - Find the middle term in the expansion of:
(i) (2/3x – 3/2x)20
(ii) (a/x + bx)12
(iii) (x2 – 2/x)10
(iv) (x/a – a/x)10
Answer - 13 : -
(i) (2/3x – 3/2x)20
We have,
(2/3x – 3/2x)20 where,n = 20 (even number)
So the middle term is(n/2 + 1) = (20/2 + 1) = (10 + 1) = 11. ie., 11th term
Now,
T11 =T10+1
= 20C10 (2/3x)20-10 (3/2x)10
= 20C10 210/310 ×310/210 x10-10
= 20C10
Hence, the middle termis 20C10.
(ii) (a/x + bx)12
We have,
(a/x + bx)12 where,n = 12 (even number)
So the middle term is(n/2 + 1) = (12/2 + 1) = (6 + 1) = 7. ie., 7th term
Now,
T7 = T6+1
= 924 a6b6
Hence, the middle termis 924 a6b6.
(iii) (x2 –2/x)10
We have,
(x2 –2/x)10 where, n = 10 (even number)
So the middle term is(n/2 + 1) = (10/2 + 1) = (5 + 1) = 6. ie., 6th term
Now,
T6 = T5+1
Hence, the middle termis -8064x5.
(iv) (x/a – a/x)10
We have,
(x/a – a/x) 10 where,n = 10 (even number)
So the middle term is(n/2 + 1) = (10/2 + 1) = (5 + 1) = 6. ie., 6th term
Now,
T6 = T5+1
Hence, the middle termis -252.
Question - 14 : - Find the middle terms in the expansion of:
(i) (3x – x3/6)9
(ii) (2x2 – 1/x)7
(iii) (3x – 2/x2)15
(iv) (x4 – 1/x3)11
Answer - 14 : -
(i) (3x – x3/6)9
We have,
(3x – x3/6)9 where,n = 9 (odd number)
So the middle termsare ((n+1)/2) = ((9+1)/2) = 10/2 = 5 and
((n+1)/2 + 1) =((9+1)/2 + 1) = (10/2 + 1) = (5 + 1) = 6
The terms are 5th and6th.
Now,
T5 = T4+1
Hence, the middle termare 189/8 x17 and -21/16 x19.
(ii) (2x2 –1/x)7
We have,
(2x2 –1/x)7 where, n = 7 (odd number)
So the middle termsare ((n+1)/2) = ((7+1)/2) = 8/2 = 4 and
((n+1)/2 + 1) =((7+1)/2 + 1) = (8/2 + 1) = (4 + 1) = 5
The terms are 4th and5th.
Now,
Hence, the middle termare -560x5 and 280x2.
(iii) (3x – 2/x2)15
We have,
(3x – 2/x2)15 where,n = 15 (odd number)
So the middle termsare ((n+1)/2) = ((15+1)/2) = 16/2 = 8 and
((n+1)/2 + 1) =((15+1)/2 + 1) = (16/2 + 1) = (8 + 1) = 9
The terms are 8th and9th.
Now,
Hence, the middle termare (-6435×38×27)/x6 and (6435×37×28)/x9.
(iv) (x4 –1/x3)11
We have,
(x4 –1/x3)11
where, n = 11 (oddnumber)
So the middle termsare ((n+1)/2) = ((11+1)/2) = 12/2 = 6 and
((n+1)/2 + 1) =((11+1)/2 + 1) = (12/2 + 1) = (6 + 1) = 7
The terms are 6th and7th.
Now,
T7 = T6+1
Hence, the middle termare -462x9 and 462x2.
Question - 15 : - Find the middle terms in the expansion of:
(i) (x – 1/x)10
(ii) (1 – 2x + x2)n
(iii) (1 + 3x + 3x2 + x3)2n
(iv) (2x – x2/4)9
(v) (x – 1/x)2n+1
(vi) (x/3 + 9y)10
(vii) (3 – x3/6)7
(viii) (2ax – b/x2)12
(ix) (p/x + x/p)9
(x) (x/a – a/x)10
Answer - 15 : -
(i) (x – 1/x)10
We have,
(x – 1/x)10 where,n = 10 (even number)
So the middle term is(n/2 + 1) = (10/2 + 1) = (5 + 1) = 6. ie., 6th term
Now,
T6 = T5+1
Hence, the middle termis -252.
(ii) (1 – 2x + x2)n
We have,
(1 – 2x + x2)n =(1 – x)2n where, n is an even number.
So the middle term is(2n/2 + 1) = (n + 1)th term.
Now,
Tn = Tn+1
= 2nCn (-1)n (x)n
= (2n)!/(n!)2 (-1)n xn
Hence, the middle termis (2n)!/(n!)2 (-1)n xn.
(iii) (1 + 3x + 3x2 +x3)2n
We have,
(1 + 3x + 3x2 +x3)2n = (1 + x)6n where, n is aneven number.
So the middle term is(n/2 + 1) = (6n/2 + 1) = (3n + 1)th term.
Now,
T2n =T3n+1
= 6nC3n x3n
= (6n)!/(3n!)2 x3n
Hence, the middle termis (6n)!/(3n!)2 x3n.
(iv) (2x – x2/4)9
We have,
(2x – x2/4)9 where,n = 9 (odd number)
So the middle termsare ((n+1)/2) = ((9+1)/2) = 10/2 = 5 and
((n+1)/2 + 1) =((9+1)/2 + 1) = (10/2 + 1) = (5 + 1) = 6
The terms are 5th and6th.
Now,
T5 = T4+1
And,
T6 = T5+1
Hence, the middle termis 63/4 x13 and -63/32 x14.
(v) (x – 1/x)2n+1
We have,
(x – 1/x)2n+1 where,n = (2n + 1) is an (odd number)
So the middle termsare ((n+1)/2) = ((2n+1+1)/2) = (2n+2)/2 = (n + 1) and
((n+1)/2 + 1) =((2n+1+1)/2 + 1) = ((2n+2)/2 + 1) = (n + 1 + 1) = (n + 2)
The terms are (n + 1)th and(n + 2)th.
Now,
Tn = Tn+1
And,
Tn+2 =Tn+1+1
Hence, the middle termis (-1)n.2n+1Cn x and (-1)n+1.2n+1Cn (1/x).
(vi) (x/3 + 9y)10
We have,
(x/3 + 9y)10 where,n = 10 is an even number.
So the middle term is(n/2 + 1) = (10/2 + 1) = (5 + 1) = 6. i.e., 6th term.
Now,
T6 = T5+1
Hence, the middle termis 61236x5y5.
(vii) (3 – x3/6)7
We have,
(3 – x3/6)7 where,n = 7 (odd number).
So the middle termsare ((n+1)/2) = ((7+1)/2) = 8/2 = 4 and
((n+1)/2 + 1) =((7+1)/2 + 1) = (8/2 + 1) = (4 + 1) = 5
The terms are 4th and5th.
Now,
T4 = T3+1
= 7C3 (3)7-3 (-x3/6)3
= -105/8 x9
And,
T5 = T4+1
= 9C4 (3)9-4 (-x3/6)4
Hence, the middleterms are -105/8 x9 and 35/48 x12.
(viii) (2ax – b/x2)12
We have,
(2ax – b/x2)12 where,n = 12 is an even number.
So the middle term is(n/2 + 1) = (12/2 + 1) = (6 + 1) = 7. i.e., 7th term.
Now,
T7 = T6+1
Hence, the middle termis (59136a6b6)/x6.
(ix) (p/x + x/p)9
We have,
(p/x + x/p)9 where,n = 9 (odd number).
So the middle termsare ((n+1)/2) = ((9+1)/2) = 10/2 = 5 and
((n+1)/2 + 1) =((9+1)/2 + 1) = (10/2 + 1) = (5 + 1) = 6
The terms are 5th and6th.
Now,
T5 = T4+1
And,
T6 = T5+1
= 9C5 (p/x)9-5 (x/p)5
Hence, the middleterms are 126p/x and 126x/p.
(x) (x/a – a/x)10
We have,
(x/a – a/x) 10 where,n = 10 (even number)
So the middle term is(n/2 + 1) = (10/2 + 1) = (5 + 1) = 6. ie., 6th term
Now,
T6 = T5+1
Hence, the middle termis -252.
Question - 16 : - Find the term independent of x in the expansion of the followingexpressions:
(i) (3/2 x2 – 1/3x)9
(ii) (2x + 1/3x2)9
(iii) (2x2 – 3/x3)25
(iv) (3x – 2/x2)15
(v) ((√x/3) + √3/2x2)10
(vi) (x – 1/x2)3n
(vii) (1/2 x1/3 + x-1/5)8
(viii) (1 + x + 2x3) (3/2x2 – 3/3x)9
(ix) (∛x + 1/2∛x)18, x> 0
(x) (3/2x2 – 1/3x)6
Answer - 16 : -
(i) (3/2 x2 –1/3x)9
Given:
(3/2 x2 –1/3x)9
If (r + 1)th termin the given expression is independent of x.
Then, we have:
Tr+1 = nCr xn-r ar
For this term to beindependent of x, we must have
18 – 3r = 0
3r = 18
r = 18/3
= 6
So, the required termis 7th term.
We have,
T7 = T6+1
= 9C6 ×(39-12)/(29-6)
= (9×8×7)/(3×2) × 3-3 ×2-3
= 7/18
Hence, the termindependent of x is 7/18.
(ii) (2x + 1/3x2)9
Given:
(2x + 1/3x2)9
If (r + 1)th termin the given expression is independent of x.
Then, we have:
Tr+1 = nCr xn-r ar
For this term to beindependent of x, we must have
9 – 3r = 0
3r = 9
r = 9/3
= 3
So, the required termis 4th term.
We have,
T4 = T3+1
= 9C3 ×(26)/(33)
= 9C3 ×64/27
Hence, the termindependent of x is 9C3 × 64/27.
(iii) (2x2 –3/x3)25
Given:
(2x2 –3/x3)25
If (r + 1)th termin the given expression is independent of x.
Then, we have:
Tr+1 = nCr xn-r ar
= 25Cr (2x2)25-r (-3/x3)r
= (-1)r 25Cr ×225-r × 3r x50-2r-3r
For this term to beindependent of x, we must have
50 – 5r = 0
5r = 50
r = 50/5
= 10
So, the required termis 11th term.
We have,
T11 =T10+1
= (-1)10 25C10 ×225-10 × 310
= 25C10 (215 ×310)
Hence, the termindependent of x is 25C10 (215 ×310).
(iv) (3x – 2/x2)15
Given:
(3x – 2/x2)15
If (r + 1)th termin the given expression is independent of x.
Then, we have:
Tr+1 = nCr xn-r ar
= 15Cr (3x)15-r (-2/x2)r
= (-1)r 15Cr ×315-r × 2r x15-r-2r
For this term to beindependent of x, we must have
15 – 3r = 0
3r = 15
r = 15/3
= 5
So, the required termis 6th term.
We have,
T6 = T5+1
= (-1)5 15C5 ×315-5 × 25
= -3003 × 310 ×25
Hence, the termindependent of x is -3003 × 310 × 25.
(v) ((√x/3) + √3/2x2)10
Given:
((√x/3) + √3/2x2)10
If (r + 1)th termin the given expression is independent of x.
Then, we have:
Tr+1 = nCr xn-r ar
For this term to beindependent of x, we must have
(10-r)/2 – 2r = 0
10 – 5r = 0
5r = 10
r = 10/5
= 2
So, the required termis 3rd term.
We have,
T3 = T2+1
Hence, the termindependent of x is 5/4.
(vi) (x – 1/x2)3n
Given:
(x – 1/x2)3n
If (r + 1)th termin the given expression is independent of x.
Then, we have:
Tr+1 = nCr xn-r ar
= 3nCr x3n-r (-1/x2)r
= (-1)r 3nCr x3n-r-2r
For this term to beindependent of x, we must have
3n – 3r = 0
r = n
So, the required termis (n+1)th term.
We have,
(-1)n 3nCn
Hence, the termindependent of x is (-1)n 3nCn
(vii) (1/2 x1/3 +x-1/5)8
Given:
(1/2 x1/3 +x-1/5)8
If (r + 1)th termin the given expression is independent of x.
Then, we have:
Tr+1 = nCr xn-r ar
For this term to beindependent of x, we must have
(8-r)/3 – r/5 = 0
(40 – 5r – 3r)/15 = 0
40 – 5r – 3r = 0
40 – 8r = 0
8r = 40
r = 40/8
= 5
So, the required termis 6th term.
We have,
T6 = T5+1
= 8C5 ×1/(28-5)
= (8×7×6)/(3×2×8)
= 7
Hence, the termindependent of x is 7.
(viii) (1 + x + 2x3)(3/2x2 – 3/3x)9
Given:
(1 + x + 2x3)(3/2x2 – 3/3x)9
If (r + 1)th termin the given expression is independent of x.
Then, we have:
(1 + x + 2x3)(3/2x2 – 3/3x)9 =
= 7/18 – 2/27
= (189 – 36)/486
= 153/486 (divide by9)
= 17/54
Hence, the termindependent of x is 17/54.
(ix) (∛x + 1/2∛x)18, x > 0
Given:
(∛x + 1/2∛x)18, x > 0
If (r + 1)th termin the given expression is independent of x.
Then, we have:
Tr+1 = nCr xn-r ar
For this term to beindependent of r, we must have
(18-r)/3 – r/3 = 0
(18 – r – r)/3 = 0
18 – 2r = 0
2r = 18
r = 18/2
= 9
So, the required termis 10th term.
We have,
T10 =T9+1
= 18C9 ×1/29
Hence, the termindependent of x is 18C9 × 1/29.
(x) (3/2x2 –1/3x)6
Given:
(3/2x2 –1/3x)6
If (r + 1)th termin the given expression is independent of x.
Then, we have:
Tr+1 = nCr xn-r ar
For this term to beindependent of r, we must have
12 – 3r = 0
3r = 12
r = 12/3
= 4
So, the required termis 5th term.
We have,
T5 = T4+1
Hence, the termindependent of x is 5/12.
Question - 17 : - If the coefficients of (2r + 4)th and (r – 2)th terms in the expansion of(1 + x)18 are equal, find r.
Answer - 17 : -
Given:
(1 + x)18
We know, thecoefficient of the r term in the expansion of (1 + x)n is nCr-1
So, the coefficientsof the (2r + 4) and (r – 2) terms in the given expansion are 18C2r+4-1 and 18Cr-2-1
For these coefficientsto be equal, we must have
18C2r+4-1 = 18Cr-2-1
18C2r+3 = 18Cr-3
2r + 3 = r – 3 (or) 2r+ 3 + r – 3 = 18 [Since, nCr = nCs =>r = s (or) r + s = n]
2r – r = -3 – 3 (or)3r = 18 – 3 + 3
r = -6 (or) 3r = 18
r = -6 (or) r = 18/3
r = -6 (or) r = 6
∴ r = 6 [since, rshould be a positive integer.]
Question - 18 : - If the coefficients of (2r + 1)th term and (r + 2)th term inthe expansion of (1 + x)43 are equal, find r.
Answer - 18 : -
Given:
(1 + x)43
We know, thecoefficient of the r term in the expansion of (1 + x)n is nCr-1
So, the coefficientsof the (2r + 1) and (r + 2) terms in the given expansion are 43C2r+1-1 and 43Cr+2-1
For these coefficientsto be equal, we must have
43C2r+1-1 = 43Cr+2-1
43C2r = 43Cr+1
2r = r + 1 (or) 2r + r+ 1 = 43 [Since, nCr = nCs =>r = s (or) r + s = n]
2r – r = 1 (or) 3r + 1= 43
r = 1 (or) 3r = 43 – 1
r = 1 (or) 3r = 42
r = 1 (or) r = 42/3
r = 1 (or) r = 14
∴ r = 14 [since, value‘1’ gives the same term]
Question - 19 : - Prove that the coefficient of (r + 1)th term in the expansion of (1+ x)n + 1 is equal to the sum of the coefficientsof rth and (r + 1)th terms in the expansion of (1 + x)n.
Answer - 19 : -
We know, thecoefficients of (r + 1)th term in (1 + x)n+1 is n+1Cr
So, sum of thecoefficients of the rth and (r + 1)th terms in (1 + x)n is
(1 + x)n = nCr-1 + nCr
= n+1Cr [since, nCr+1 + nCr = n+1Cr+1]
Hence proved.