# Class 11 RD Sharma Solutions- Chapter 18 Binomial Theorem – Exercise 18.1

### (i) (2x + 3y)5

Solution:

Using binomial theorem, we have,

(2x + 3y)5 = 5C0 (2x)5 (3y)0 + 5C1 (2x)4 (3y)1 + 5C2 (2x)3 (3y)2 + 5C3 (2x)2 (3y)3 + 5C4 (2x)1 (3y)4 + 5C5 (2x)0 (3y)5

= 32x5 + 5 (16x4) (3y) + 10 (8x3) (9y)2 + 10 (4x)2 (27y)3 + 5 (2x) (81y4) + 243 y5

= 32x5 + 240x4 y + 720x3y2 + 1080x2y3 + 810xy4 + 243y5

### (ii) (2x â€“ 3y)4

Solution:

Using binomial theorem, we have,

(2x â€“ 3y)4 = 4C0 (2x)4 (3y)0 â€“ 4C1 (2x)3 (3y)1 + 4C2 (2x)2 (3y)2 â€“ 4C3 (2x)1 (3y)3 + 4C4 (2x)0 (3y)4

= 16x4 â€“ 4 (8x3) (3y) + 6 (4x2) (9y2) â€“ 4 (2x) (27y3) + 81y4

= 16x4 â€“ 96x3y + 216x2y2 â€“ 216xy3 + 81y4

### (iii) (x – 1/x)6

Solution:

Using binomial theorem, we have,

(x – 1/x)6 = 6C0 x6 (1/x)06C1 x5 (1/x)1 + 6C2 x4 (1/x)26C3 x3 (1/x)3 + 6C4 x2 (1/x)46C5 x1 (1/x)5 + 6C6 (1/x)6

= x6 – 6x5 (1/x) + 15x4 (1/x2) – 20 x3 (1/x3) + 15x2 (1/x4) – 6x (1/x5) + 1/x6

= x6 – 6x4 + 15x2 – 20 + 15/x2 – 6/x4 + 1/x6

### (iv) (1 â€“ 3x)7

Solution:

Using binomial theorem, we have,

(1 â€“ 3x)7 = 7C0 (3x)0 â€“ 7C1 (3x)1 + 7C2 (3x)2 â€“ 7C3 (3x)3 + 7C4 (3x)4 â€“ 7C5 (3x)5 + 7C6 (3x)6 â€“ 7C7 (3x)7

= 1 â€“ 7 (3x) + 21 (9x)2 â€“ 35 (27x3) + 35 (81x4) â€“ 21 (243x5) + 7 (729x6) â€“ 2187(x7)

= 1 â€“ 21x + 189x2 â€“ 945x3 + 2835x4 â€“ 5103x5 + 5103x6 â€“ 2187x7

### (v) (ax – b/x)6

Solution:

Using binomial theorem, we have,

(ax – b/x)^6 =

### (vi)

Solution:

Using binomial theorem, we have,

### (vii)

Solution:

Using binomial theorem, we have,

### (viii) (1 + 2x â€“ 3x2)5

Solution:

Using binomial theorem, we have,

(1 + 2x â€“ 3x2)5 = 5C0 (1 + 2x)5 (3x2)0 â€“ 5C1 (1 + 2x)4 (3x2)1 + 5C2 (1 + 2x)3 (3x2)2 â€“ 5C3 (1 + 2x)2 (3x2)3 + 5C4 (1 + 2x)1 (3x2)4 â€“ 5C5 (1 + 2x)0 (3x2)5

= (1 + 2x)5 â€“ 5(1 + 2x)4 (3x2) + 10 (1 + 2x)3 (9x4) â€“ 10 (1 + 2x)2 (27x6) + 5 (1 + 2x) (81x8) â€“ 243x10

= 5C0 (2x)0 + 5C1 (2x)1 + 5C2 (2x)2 + 5C3 (2x)3 + 5C4 (2x)4 + 5C5 (2x)5 â€“ 15x2 [4C0 (2x)0 + 4C1 (2x)1 + 4C2 (2x)2 + 4C3 (2x)3 + 4C4 (2x)4] + 90x4 [1 + 8x3 + 6x + 12x2] â€“ 270x6(1 + 4x2 + 4x) + 405x8 + 810x9 â€“ 243x10

= 1 + 10x + 40x2 + 80x3 + 80x4 + 32x5 â€“ 15x2 â€“ 120x3 â€“ 3604 â€“ 480x5 â€“ 240x6 + 90x4 + 720x7 + 540x5 + 1080x6 â€“ 270x6 â€“ 1080x8 â€“ 1080x7 + 405x8 + 810x9 â€“ 243x10

= 1 + 10x + 25x2 â€“ 40x3 â€“ 190x4 + 92x5 + 570x6 â€“ 360x7 â€“ 675x8 + 810x9 â€“ 243x10

### (ix)

Solution:

Using binomial theorem, we have,

(x + 1 – 1/x)3 = 3C0 (x + 1)3 (1/x)0 3C1(x + 1)2(1/x)1 + 3C2(x + 1)1(1/x)23C3 (x + 1)0 (1/x)3

### (x) (1 – 2x + 3x2)3

Solution:

Using binomial theorem, we have,

(1 – 2x + 3 x2)3 = 3C0 (1 – 2x)3 + 3C1 (1 – 2x)2 (3x2) + 3C2 (1 – 2x)(3x2)2 + 3C3 (3x2)3

= (1 – 2x)3 + 9x2 (1 – 2x)2 + 27x4 (1 – 2x) + 27x6

= 1 – 8x3 + 12x2 – 6x + 9x2 (1 + 4x2 – 4x) + 27x4 – 54x5 + 27x

= 1 – 8x3 + 12x2 – 6x + 9x2 + 36x4 – 36x3 + 27x4 – 54x5 + 27x6

= 1 – 6x + 21x2 – 44x3 + 63x4 – 54x5 + 27x

### (i)

Solution:

Using binomial theorem, we have,

=

= 2 [(x + 1)3 + 15(x + 1)2 (x – 1) + 15(x + 1)(x – 1 )2 + (x – 1 )3]

= 2 [x3 + 1 + 3x + 3 x2 + 15( x2 + 2x + 1)(x – 1) + 15(x + 1)( x2 + 1 – 2x) + x3 – 1 + 3x – 3 x2]

= 2 [2 x3 + 6x + 15 x3 – 15 x2 + 30 x2 – 30x + 15x – 15 + 15 x3 + 15 x2 – 30 x2 – 30x + 15x + 15]

= 2 [32 x3 – 24x]

= 16x [4x2 – 3]

### (ii)

Solution:

Using binomial theorem, we have,

= 2 [x6 + 15 x4 ( x2 – 1) + 15 x2 ( x2 – 1 )2 + ( x2 – 1 )3]

= 2 [x6 + 15 x6 – 15 x4 + 15 x2 ( x4 – 2 x2 + 1) + ( x6 – 1 + 3 x2 – 3 x4)]

= 2 [x6 + 15 x6 – 15 x4 + 15 x6 – 30 x4 + 15 x2 + x6 – 1 + 3 x2 – 3 x4]

= 64 x6 – 96 x4 + 36 x2 – 2

### (iii)

Solution:

Using binomial theorem, we have,

= 2 [5C0 (2âˆšx)0 + 5C2 (2âˆšx)2 + 5C4 (2âˆšx)4]

= 2 [1 + 10 (4x) + 5 (16x2)]

= 2 [1 + 40x + 80x2]

### (iv)

Solution:

Using binomial theorem, we have,

= 2 [6C0 (âˆš2)6 + 6C2 (âˆš2)4 + 6C4 (âˆš2)2 + 6C6 (âˆš2)0]

= 2 [8 + 15 (4) + 15 (2) + 1]

= 2 [99]

= 198

### (v)

Solution:

Using binomial theorem, we have,

= 2 [5C1 (34) (âˆš2)1 + 5C3 (32) (âˆš2)3 + 5C5 (30) (âˆš2)5]

= 2 [5 (81) (âˆš2) + 10 (9) (2âˆš2) + 4âˆš2]

= 2âˆš2 (405 + 180 + 4)

= 1178âˆš2

### (vi)

Solution:

Using binomial theorem, we have,

= 2 [7C0 (27) (âˆš3)0 + 7C2 (25) (âˆš3)2 + 7C4 (23) (âˆš3)4 + 7C6 (21) (âˆš3)6]

= 2 [128 + 21 (32)(3) + 35(8)(9) + 7(2)(27)]

= 2 [128 + 2016 + 2520 + 378]

= 2 [5042]

= 10084

### (vii)

Solution:

Using binomial theorem, we have,

= 2 [5C1 (âˆš3)4 + 5C3 (âˆš3)2 + 5C5 (âˆš3)0]

= 2 [5 (9) + 10 (3) + 1]

= 2 [76]

= 152

### (viii) (0.99)5 + (1.01)5

Solution:

Using binomial theorem, we have,

(0.99)5 + (1.01)5 = (1 â€“ 0.01)5 + (1 + 0.01)5

= 2 [5C0 (0.01)0 + 5C2 (0.01)2 + 5C4 (0.01)4]

= 2 [1 + 10 (0.0001) + 5 (0.00000001)]

= 2 [1.00100005]

= 2.0020001

### (ix)

Solution:

Using binomial theorem, we have,

= 2 [6C1 (âˆš3)5 (âˆš2)1 + 6C3 (âˆš3)3 (âˆš2)3 + 6C5 (âˆš3)1 (âˆš2)5]

= 2 [6 (9âˆš3) (âˆš2) + 20 (3âˆš3) (2âˆš2) + 6 (âˆš3) (4âˆš2)]

= 2 [âˆš6 (54 + 120 + 24)]

= 396 âˆš6

(x)

Solution:

Using binomial theorem, we have,

= 2[ a8 + 6 a4 (a2 – 1) + ( a2 – 1 )2]

= 2[a8 + 6 a6 – 6 a4 + a4 + 1 – 2 a2]

= 2 a8 + 12 a6 – 10 a4 – 4 a2 + 2

### Question 3. Find (a + b)4 â€“ (a â€“ b)4. Hence, evaluate (âˆš3 + âˆš2)4 â€“ (âˆš3 â€“ âˆš2)4.

Solution:

We are given,

(a + b)4 â€“ (a â€“ b)4 = 2 [4C1 a3b1 + 4C3 a1b3]

= 2 [4a3b + 4ab3]

= 8 (a3b + ab3)

Now,

(âˆš3 + âˆš2)4 â€“ (âˆš3 â€“ âˆš2)4 = 8 (a3b + ab3)

= 8 [(âˆš3)3 (âˆš2) + (âˆš3) (âˆš2)3]

= 8 [(3âˆš6) + (2âˆš6)]

= 8 (5âˆš6)

= 40âˆš6

### Question 4. Find (x + 1)6 + (x â€“ 1)6. Hence, or otherwise evaluate (âˆš2 + 1)6 + (âˆš2 â€“ 1)6.

Solution:

We are given,

(x + 1)6 + (x â€“ 1)6 = 2 [6C0 x6 + 6C2 x4 + 6C4 x2 + 6C6 x0]

= 2 [x6 + 15x4 + 15x2 + 1]

Now,

(âˆš2 + 1)6 + (âˆš2 â€“ 1)6

So consider, x = âˆš2 then we get,

(âˆš2 + 1)6 + (âˆš2 â€“ 1)6 = 2 [x6 + 15x4 + 15x2 + 1]

= 2 [(âˆš2)6 + 15 (âˆš2)4 + 15 (âˆš2)2 + 1]

= 2 [8 + 15 (4) + 15 (2) + 1]

= 2 [8 + 60 + 30 + 1]

= 198

### (i) (96)3

Solution:

On expressing the given expression as two different entities and applying the binomial theorem, we get,

(96)3 = (100 â€“ 4)3

= 3C0 (100)3 (4)0 â€“ 3C1 (100)2 (4)1 + 3C2 (100)1 (4)2 â€“ 3C3 (100)0 (4)3

= 1000000 â€“ 120000 + 4800 â€“ 64

= 884736

### (ii) (102)5

Solution:

On expressing the given expression as two different entities and applying the binomial theorem, we get,

(102)5 = (100 + 2)5

= 5C0 (100)5 (2)0 + 5C1 (100)4 (2)1 + 5C2 (100)3 (2)2 + 5C3 (100)2 (2)3 + 5C4 (100)1 (2)4 + 5C5 (100)0 (2)5

= 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32

= 11040808032

### (iii) (101)4

Solution:

On expressing the given expression as two different entities and applying the binomial theorem, we get,

(101)4 = (100 + 1)4

= 4C0 (100)4 + 4C1 (100)3 + 4C2 (100)2 + 4C3 (100)1 + 4C4 (100)0

= 100000000 + 4000000 + 60000 + 400 + 1

= 104060401

### (iv) (98)5

Solution:

On expressing the given expression as two different entities and applying the binomial theorem, we get,

(98)5 = (100 â€“ 2)5

= 5C0 (100)5 (2)0 â€“ 5C1 (100)4 (2)1 + 5C2 (100)3 (2)2 â€“ 5C3 (100)2 (2)3 + 5C4 (100)1 (2)4 â€“ 5C5 (100)0 (2)5

= 10000000000 â€“ 1000000000 + 40000000 â€“ 800000 + 8000 â€“ 32

= 9039207968

### Question 6. Using binomial theorem, prove that 23n â€“ 7n â€“ 1 is divisible by 49, where n âˆˆ N.

Solution:

We are given,

23n â€“ 7n â€“ 1 = 8n â€“ 7n â€“ 1

= (1 + 7)n â€“ 7n â€“ 1

= nC0 + nC1 (7)1 + nC2 (7)2 + nC3 (7)3 + nC4 (7)2 + nC5 (7)1 + .â€¦ + nCn (7)n  â€“ 7n â€“ 1

= 1 + 7n + 72 [ nC2 + nC3 (71) + nC4 (72) + â€¦ + nCn (7)n-2] â€“ 7n â€“ 1

= 49 [ nC2 + nC3 (71) + nC4 (72) + â€¦ + nCn (7)n-2] , which is divisible by 49.

Therefore, 23n â€“ 1 â€“ 7n is divisible by 49.

Hence proved.

### Question 7. Using the binomial theorem, prove that 32n+2 â€“ 8n â€“ 9 is divisible by 64, where n âˆˆ N.

Solution:

We are given,

32n+2 â€“ 8n â€“ 9 = 32(n+1) â€“ 8n â€“ 9

= 9n+1 â€“ 8n â€“ 9

= (1 + 8)n+1 â€“ 8n â€“ 9

= n+1C0 + n+1C1 (8)1 + n+1C2 (8)2 + n+1C3 (8)3 + n+1C4 (8)2 + n+1C5 (8)1 + .â€¦ + n+1Cn+1 (8)n+1 â€“ 8n â€“ 9

= 1 + 8(n+1) + 82 [ n+1C2 + n+1C3 (81) + n+1C4 (82) + â€¦ + n+1Cn+1 (8)n-1] â€“ 8n â€“ 9

= 8n + 9 + 64 [ n+1C2 + n+1C3 (81) + n+1C4 (82) + â€¦ + n+1Cn+1 (8)n-1] â€“ 8n â€“ 9

= 64 [ n+1C2 + n+1C3 (81) + n+1C4 (82) + â€¦ + n+1Cn+1 (8)n-1], which is divisible by 64.

Therefore, 32n+2 â€“ 8n â€“ 9 is divisible by 64.

Hence proved.

### Question 8. If n is a positive integer, prove that 33n â€“ 26n â€“ 1 is divisible by 676.

Solution:

We are given,

33n â€“ 26n â€“ 1 = (33)n â€“ 26n â€“ 1

= 27n â€“ 26n â€“ 1

= (1 + 26)n â€“ 26n â€“ 1

= nC0 + nC1 (26)1 + nC2 (26)2 + nC3 (26)3 + nC4 (26)2 + nC5 (26)1 + .â€¦ + nCn (26)n  â€“ 26n â€“ 1

= 1 + 26n + 262 [ nC2 + nC3 (261) + nC4 (262) + â€¦ + nCn (26)n-2] â€“ 26n â€“ 1

= 676 [ nC2 + nC3 (261) + nC4 (262) + â€¦ + nCn (26)n-2] , which is divisible by 676.

Therefore, 33n â€“ 26n â€“ 1 is divisible by 676.

Hence proved.

### Question 9. Using binomial theorem, indicate which is larger (1.1)10000 or 1000.

Solution:

We have,

(1.1)10000 = (1 + 0.1)10000

= 10000C0 + 10000C1 (0.1)1 + 10000C2 (0.1)2 + .â€¦ + 10000C10000 (0.1)10000

= 1 + (10000) (0.1) + other positive terms

= 1 + 1000 + other positive terms

= 1001 + other positive terms

Therefore, (1.1)10000 is larger than 1000.

### Question 10. Using binomial theorem, determine which number is larger, (1.2)4000 or 800?

Solution:

We have,

(1.2)4000 = (1 + 0.2)4000

= 4000C0 + 4000C1 (0.2)1 + 4000C2 (0.2)2 + .â€¦ + 4000C4000 (0.2)4000

= 1 + (4000) (0.2) + other positive terms

= 1 + 800 + other positive terms

= 801 + other positive terms

Therefore, (1.2)4000 is larger than 800.

### Question 11. Find the value of (1.01)10 + (1âˆ’0.01)10 correct to 7 places of decimal.

Solution:

We have,

(1.01)10 + (1âˆ’0.01)10 = (1+0.01)10 + (1âˆ’0.01)10

= 2.0090042

### Question 12. Show that 24n+4 âˆ’ 15n âˆ’ 16, where n âˆˆ N is divisible by 225.

Solution:

We have,

24n+4 âˆ’ 15n âˆ’ 16 = 24(n+1) âˆ’ 15n âˆ’ 16

= 16n+1 âˆ’ 15n âˆ’ 16

= (1 + 15)n+1 âˆ’ 15n âˆ’ 16

= n+1C0 + n+1C1 (15)1 + n+1C2 (15)2 + n+1C3 (15)3 + n+1C4 (15)2 + n+1C5 (15)1 + .â€¦ + n+1Cn+1 (15)n+1 âˆ’ 15n âˆ’ 16

= 1 + 15(n+1) + 152 [ n+1C2 + n+1C3 (151) + n+1C4 (152) + â€¦ + n+1Cn+1 (15)n-1] â€“ 15n â€“ 16

= 15n + 16 + 225 [ n+1C2 + n+1C3 (151) + n+1C4 (152) + â€¦ + n+1Cn+1 (15)n-1] â€“ 15n â€“ 16

= 225 [ n+1C2 + n+1C3 (151) + n+1C4 (152) + â€¦ + n+1Cn+1 (15)n-1] , which is divisible by 225.

Therefore, 24n+4 âˆ’ 15n âˆ’ 16 is divisible by 225.

Hence proved.

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