RD Chapter 18 Binomial Theorem Ex 18.1 Solutions
Question - 1 : - Using binomial theorem, write down the expressions of the following:
Answer - 1 : -
(i) (2x + 3y) 5
Let us solve the givenexpression:
(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) + 243y5
= 32x5 +240x4y + 720x3y2 + 1080x2y3 +810xy4 + 243y5
(ii) (2x – 3y) 4
Let us solve the givenexpression:
(2x – 3y) 4 = 4C0 (2x)4 (3y)0 – 4C1 (2x)3 (3y)1 + 4C2 (2x)2 (3y)2 – 4C3 (2x)1 (3y)3 + 4C4 (2x)0 (3y)4
= 16x4 –4 (8x3) (3y) + 6 (4x2) (9y2) – 4 (2x) (27y3)+ 81y4
= 16x4 –96x3y + 216x2y2 – 216xy3 +81y4
(iv) (1 – 3x) 7
Let us solve the givenexpression:
(1 – 3x) 7 = 7C0 (3x)0 – 7C1 (3x)1 + 7C2 (3x)2 – 7C3 (3x)3 + 7C4 (3x)4 – 7C5 (3x)5 + 7C6 (3x)6 – 7C7 (3x)7
= 1 – 7 (3x) + 21 (9x)2 –35 (27x3) + 35 (81x4) – 21 (243x5) + 7 (729x6)– 2187(x7)
= 1 – 21x + 189x2 –945x3 + 2835x4 – 5103x5 + 5103x6 –2187x7
(viii) (1 + 2x – 3x2)5
Let us solve the givenexpression:
Let us consider (1 +2x) and 3x2 as two different entities and apply the binomialtheorem.
(1 + 2x – 3x2)5 = 5C0 (1+ 2x)5 (3x2)0 – 5C1 (1+ 2x)4 (3x2)1 + 5C2 (1+ 2x)3 (3x2)2 – 5C3 (1+ 2x)2 (3x2)3 + 5C4 (1+ 2x)1 (3x2)4 – 5C5 (1+ 2x)0 (3x2)5
= (1 + 2x)5 –5(1 + 2x)4 (3x2) + 10 (1 + 2x)3 (9x4)– 10 (1 + 2x)2 (27x6) + 5 (1 + 2x) (81x8)– 243x10
= 5C0 (2x)0 + 5C1 (2x)1 + 5C2 (2x)2 + 5C3 (2x)3 + 5C4 (2x)4 + 5C5 (2x)5 –15x2 [4C0 (2x)0 + 4C1 (2x)1 + 4C2 (2x)2 + 4C3 (2x)3 + 4C4 (2x)4]+ 90x4 [1 + 8x3 + 6x + 12x2] – 270x6(1+ 4x2 + 4x) + 405x8 + 810x9 –243x10
= 1 + 10x + 40x2 +80x3 + 80x4 + 32x5 – 15x2 –120x3 – 3604 – 480x5 – 240x6 +90x4 + 720x7 + 540x5 + 1080x6 –270x6 – 1080x8 – 1080x7 + 405x8 +810x9 – 243x10
= 1 + 10x + 25x2 –40x3 – 190x4 + 92x5 + 570x6 –360x7 – 675x8 + 810x9 – 243x10
(x) (1 – 2x + 3x2)3
Let us solve the givenexpression:
Question - 2 : - Evaluate the following:
Answer - 2 : -
Let us solve the givenexpression:
= 2 [5C0 (2√x)0 + 5C2 (2√x)2 + 5C4 (2√x)4]
= 2 [1 + 10 (4x) + 5(16x2)]
= 2 [1 + 40x + 80x2]
Let us solve the givenexpression:
= 2 [6C0 (√2)6 + 6C2 (√2)4 + 6C4 (√2)2 + 6C6 (√2)0]
= 2 [8 + 15 (4) + 15(2) + 1]
= 2 [99]
= 198
Let us solve the givenexpression:
= 2 [5C1 (34)(√2)1 + 5C3 (32)(√2)3 + 5C5 (30)(√2)5]
= 2 [5 (81) (√2) + 10 (9) (2√2) + 4√2]
= 2√2 (405 + 180 + 4)
= 1178√2
Let us solve the givenexpression:
= 2 [7C0 (27)(√3)0 + 7C2 (25)(√3)2 + 7C4 (23)(√3)4 + 7C6 (21)(√3)6]
= 2 [128 + 21 (32)(3)+ 35(8)(9) + 7(2)(27)]
= 2 [128 + 2016 + 2520+ 378]
= 2 [5042]
= 10084
Let us solve the givenexpression:
= 2 [5C1 (√3)4 + 5C3 (√3)2 + 5C5 (√3)0]
= 2 [5 (9) + 10 (3) +1]
= 2 [76]
= 152
Let us solve the givenexpression:
= (1 – 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
Let us solve the givenexpression:
= 2 [6C1 (√3)5 (√2)1 + 6C3 (√3)3 (√2)3 + 6C5 (√3)1 (√2)5]
= 2 [6 (9√3) (√2) + 20 (3√3) (2√2) + 6 (√3) (4√2)]
= 2 [√6 (54 + 120 + 24)]
= 396 √6
= 2 [a8 +6a6 – 6a4 + a4 + 1 – 2a2]
= 2a8 +12a6 – 10a4 – 4a2 + 2
Find (a + b) 4 – (a – b) 4. Hence,evaluate (√3 + √2)4 – (√3 – √2)4.
Answer - 3 : -
Firstly, let us solvethe given expression:
(a + b) 4 –(a – b) 4
The above expressioncan be expressed as,
(a + b) 4 –(a – b) 4 = 2 [4C1 a3b1 + 4C3 a1b3]
= 2 [4a3b +4ab3]
= 8 (a3b +ab3)
Now,
Let us evaluate theexpression:
(√3 + √2)4 –(√3 -√2)4
So consider, a = √3and b = √2 we get,
(√3 + √2)4 –(√3 -√2)4 = 8 (a3b + ab3)
= 8 [(√3)3 (√2)+ (√3) (√2)3]
= 8 [(3√6) + (2√6)]
= 8 (5√6)
= 40√6
Question - 4 : - Find (x + 1) 6 + (x – 1) 6. Hence,or otherwise evaluate (√2 + 1)6 + (√2 – 1)6.
Answer - 4 : -
Firstly, let us solvethe given expression:
(x + 1) 6 +(x – 1) 6
The above expressioncan be expressed as,
(x + 1) 6 +(x – 1) 6 = 2 [6C0 x6 + 6C2 x4 + 6C4 x2 + 6C6 x0]
= 2 [x6 +15x4 + 15x2 + 1]
Now,
Let us evaluate theexpression:
(√2 + 1)6 +(√2 – 1)6
So consider, x = √2then we get,
(√2 + 1)6 +(√2 – 1)6 = 2 [x6 + 15x4 + 15x2 +1]
= 2 [(√2)6 +15 (√2)4 + 15 (√2)2 + 1]
= 2 [8 + 15 (4) + 15(2) + 1]
= 2 [8 + 60 + 30 + 1]
= 198
Question - 5 : - Using binomial theorem evaluate each of the following:
(i) (96)3
(ii) (102)5
(iii) (101)4
(iv) (98)5
Answer - 5 : -
(i) (96)3
We have,
(96)3
Let us express thegiven expression as two different entities and apply the binomial theorem.
(96)3 =(100 – 4)3
= 3C0 (100)3 (4)0 – 3C1 (100)2 (4)1 + 3C2 (100)1 (4)2 – 3C3 (100)0 (4)3
= 1000000 – 120000 +4800 – 64
= 884736
(ii) (102)5
We have,
(102)5
Let us express thegiven expression as two different entities and apply the binomial theorem.
(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
We have,
(101)4
Let us express thegiven expression as two different entities and apply the binomial theorem.
(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
We have,
(98)5
Let us express thegiven expression as two different entities and apply the binomial theorem.
(98)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
= 9039207968
Question - 6 : - Using binomial theorem, prove that 23n – 7n – 1 isdivisible by 49, where n ∈ N.
Answer - 6 : -
Given:
23n –7n – 1
So, 23n –7n – 1 = 8n – 7n – 1
Now,
8n –7n – 1
8n =7n + 1
= (1 + 7) n
= nC0 + nC1 (7)1 + nC2 (7)2 + nC3 (7)3 + nC4 (7)2 + nC5 (7)1 +… + nCn (7) n
8n = 1+ 7n + 49 [nC2 + nC3 (71)+ nC4 (72) + … + nCn (7) n-2]
8n – 1– 7n = 49 (integer)
So now,
8n – 1– 7n is divisible by 49
Or
23n –1 – 7n is divisible by 49.
Hence proved.