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Learn/ 11th Class Mathematics/ Unit 8 /Numericals

Unit 8: Mathematical Inductions and Binomial Theorem — Exercise 8 2

11th Class Mathematics · Unit 8: Mathematical Inductions and Binomial Theorem

1.Using the binomial theorem, expand: (i) (x/2 - 2/x^2)^6 (ii) (2a - x/a)^7 (iii) (sqrt(a/x) - sqrt(x/a))^6
Formula
(i) Binomial theorem (a+b)6=sumr=06binom{6}{r}a6-rbr,quad a=x2, b=-2x2
Working
Expanding =binom{6}{0}left(x2right)6left(-2x2right)0+binom{6}{1}left(x2right)5left(-2x2right)1+binom{6}{2}left(x2right)4left(-2x2right)2+binom{6}{3}left(x2right)3left(-2x2right)3+binom{6}{4}left(x2right)2left(-2x2right)4+binom{6}{5}left(x2right)1left(-2x2right)5+binom{6}{6}left(x2right)0left(-2x2right)6
=x664+6left(x532right)left(-2x2right)+15left(x416right)left(4x4right)+20left(x38right)left(-8x6right)+15left(x24right)left(16x8right)+6left(x2right)left(frac{-32}{x10right)+frac{64}{x12
Result
Expansion =x664-38x3+154-20x3+60x6-96x9+frac{64}{x12
Formula
(ii) Binomial theorem (p+q)7=sumr=07binom{7}{r}p7-rqr,quad p=2a, q=-xa
Working
Expanding =binom{7}{0}(2a)7left(-xaright)0+binom{7}{1}(2a)6left(-xaright)1+binom{7}{2}(2a)5left(-xaright)2+binom{7}{3}(2a)4left(-xaright)3+binom{7}{4}(2a)3left(-xaright)4+binom{7}{5}(2a)2left(-xaright)5+binom{7}{6}(2a)1left(-xaright)6+binom{7}{7}(2a)0left(-xaright)7
=128a7+7(64a6)left(-xaright)+21(32a5)left(x2a2right)+35(16a4)left(-x3a3right)+35(8a3)left(x4a4right)+21(4a2)left(-x5a5right)+7(2a)left(x6a6right)+1left(-x7a7right)
Result
Expansion =128a7-448a5x+672a3x2-560ax3+280x4a-84x5a3+14x6a5-x7a7
Formula
(iii) Binomial theorem (p+q)6=sumr=06binom{6}{r}p6-rqr,quad p=sqrt{ax, q=-sqrt{xa
Working
Expanding =binom{6}{0}left(sqrt{axright)6left(-sqrt{xaright)0+binom{6}{1}left(sqrt{axright)5left(-sqrt{xaright)1+binom{6}{2}left(sqrt{axright)4left(-sqrt{xaright)2+binom{6}{3}left(sqrt{axright)3left(-sqrt{xaright)3+binom{6}{4}left(sqrt{axright)2left(-sqrt{xaright)4+binom{6}{5}left(sqrt{axright)1left(-sqrt{xaright)5+binom{6}{6}left(sqrt{axright)0left(-sqrt{xaright)6
=left(axright)3+6left(axright)5/2left(-xaright)1/2+15left(axright)2left(xaright)1+20left(axright)3/2left(-xaright)3/2+15left(axright)left(xaright)2+6left(axright)1/2left(-xaright)5/2+left(xaright)3
Result
Expansion =a3x3-6a2x2+15ax-20+15xa-6x2a2+x3a3
2.Calculate by means of the binomial theorem: (i) (0.97)^3 (ii) (2.02)^4 (iii) (9.98)^4 (iv) (2.1)^5
Given
(i) Rewrite (0.97)3=(1-0.03)3
Formula
Binomial expansion =binom{3}{0}(1)3(-0.03)0+binom{3}{1}(1)2(-0.03)1+binom{3}{2}(1)1(-0.03)2+binom{3}{3}(1)0(-0.03)3
=1-0.09+0.0027-0.000027
Result
Value =0.91267
Given
(ii) Rewrite (2.02)4=(2+0.02)4
Formula
Binomial expansion =binom{4}{0}(2)4(0.02)0+binom{4}{1}(2)3(0.02)1+binom{4}{2}(2)2(0.02)2+binom{4}{3}(2)1(0.02)3+binom{4}{4}(2)0(0.02)4
=1 × 16+4(8)(0.02)+6(4)(0.0004)+4(2)(0.000008)+1(0.00000016)
=16+0.64+0.0096+0.000064+0.00000016
Result
Value =16.64966416
Given
(iii) Rewrite (9.98)4=(10-0.02)4
Formula
Binomial expansion =binom{4}{0}(10)4(-0.02)0+binom{4}{1}(10)3(-0.02)1+binom{4}{2}(10)2(-0.02)2+binom{4}{3}(10)1(-0.02)3+binom{4}{4}(10)0(-0.02)4
=1 × 10000+4(1000)(-0.02)+6(100)(0.0004)-4(10)(0.000008)+1 × (0.00000016)
=10000-80+0.24-0.00032+0.00000016
Result
Value =9920.23968016
Given
(iv) Rewrite (2.1)5=(2+0.1)5
Formula
Binomial expansion =binom{5}{0}(2)5(0.1)0+binom{5}{1}(2)4(0.1)1+binom{5}{2}(2)3(0.1)2+binom{5}{3}(2)2(0.1)3+binom{5}{4}(2)1(0.1)4+binom{5}{5}(2)0(0.1)5
=1(32)(1)+5(16)(0.1)+10(8)(0.01)+10(4)(0.001)+5(2)(0.0001)+1(0.00001)
=32+8+0.8+0.04+0.001+0.00001
Result
Value =40.84101
3.Expand and simplify: (i) (a + sqrt(2)x)^4 + (a - sqrt(2)x)^4 (ii) (2 + sqrt(3))^5 + (2 - sqrt(3))^5
Formula
(i) First expansion (a+sqrt{2}x)4=binom{4}{0}a4(sqrt{2}x)0+binom{4}{1}a3(sqrt{2}x)1+binom{4}{2}a2(sqrt{2}x)2+binom{4}{3}a(sqrt{2}x)3+binom{4}{4}a0(sqrt{2}x)4
Second expansion (a-sqrt{2}x)4=binom{4}{0}a4(sqrt{2}x)0-binom{4}{1}a3(sqrt{2}x)1+binom{4}{2}a2(sqrt{2}x)2-binom{4}{3}a(sqrt{2}x)3+binom{4}{4}a0(sqrt{2}x)4
Adding (a+sqrt{2}x)4+(a-sqrt{2}x)4=2left[binom{4}{0}a4(sqrt{2}x)0+binom{4}{2}a2(sqrt{2}x)2+binom{4}{4}a0(4x4)right]
Result
Sum =2a4+24a2x2+8x4
Formula
(ii) First expansion (2+sqrt{3})5=binom{5}{0}(2)5(sqrt{3})0+binom{5}{1}(2)4(sqrt{3})1+binom{5}{2}(2)3(sqrt{3})2+binom{5}{3}(2)2(sqrt{3})3+binom{5}{4}(2)1(sqrt{3})4+binom{5}{5}(2)0(sqrt{3})5
Second expansion (2-sqrt{3})5=binom{5}{0}(2)5(sqrt{3})0-binom{5}{1}(2)4(sqrt{3})1+binom{5}{2}(2)3(sqrt{3})2-binom{5}{3}(2)2(sqrt{3})3+binom{5}{4}(2)1(sqrt{3})4-binom{5}{5}(2)0(sqrt{3})5
Adding (2+sqrt{3})5+(2-sqrt{3})5=2left[binom{5}{0}(2)5(3)0+binom{5}{2}(2)3(sqrt{3})2+binom{5}{4}(2)1(sqrt{3})4right]=2(32+10 × 8 × 3+5 × 2 × 9)
Result
Sum =64+480+180=724
4.(i) Expand (2 + x - x^2)^4 in ascending powers of x. (ii) Expand (1 - x + x^2)^4 in ascending powers of x.
Given
(i) Substitution Let a=2+x, then (a-x2)4=sumr=04binom{4}{r}a4-r(-x2)r
(a-x2)4=a4-4a3x2+6a2x4-4ax6+x8
Working
Put a = 2+x (2+x-x2)4=(2+x)4-4x2(2+x)3+6x4(2+x)2-4x6(2+x)+x8
=(16+32x+24x2+8x3+x4)-4x2(8+12x+6x2+x3)+6x4(4+4x+x2)-4x6(2+x)+x8
Result
Expansion =16+32x-8x2-40x3+x4+20x5-2x6-4x7+x8
Given
(ii) Substitution Let a=1-x, then (1-x+x2)4=(a+x2)4=sumr=04binom{4}{r}a4-r(x2)r
(a+x2)4=a4+4a3x2+6a2x4+4ax6+x8
Working
Put a = 1-x =(1-x)4+4x2(1-x)3+6x4(1-x)2+4x6(1-x)+x8
={1-4x+6x2-4x3+x4}+4x2{1-3x+3x2-x3}+6x4(1-2x+x2)+4x6(1-x)+x8
Result
Expansion =1-4x+10x2-16x3+19x4-16x5+10x6-4x7+x8
5.(i) Find the term involving x^4 in the expansion of (3 - 2x)^7. (ii) Find the term involving x^-2 in the expansion of (x - 2/x^2)^13. (iii) Find the term involving a^4 in the expansion of (2/x - a)^9. (iv) Find the term involving y^3 in the expansion of (x - sqrt(y))^11.
Formula
(i) General term Tr+1=binom{n}{r}an-rbr,quad n=7, a=3, b=-2x
Working
Tr+1=binom{7}{r}(3)7-r(-2x)r
Set exponent of x to 4 r=4 ⇒ T5=binom{7}{4}(3)3(-2)4x4=35(27)(16)x4
Result
Required term T5=15120x4
Formula
(ii) General term Tr+1=binom{n}{r}an-rbr,quad n=13, a=x, b=-2x2
Working
Tr+1=binom{13}{r}x13-r(-2)rx-2r=binom{13}{r}(-2)r x13-3r
Set exponent of x to -2 13-3r=-2 ⇒ 15=3r ⇒ r=5
Result
Required term T5+1=binom{13}{5}(-2)5x-2=(1287)(-32)x-2=-41184x-2
Formula
(iii) General term Tr+1=binom{n}{r}an-rbr,quad n=9, abase=2x, b=-a
Working
Tr+1=binom{9}{r}left(2xright)9-r(-a)r
Set exponent of a to 4 r=4 ⇒ T4+1=binom{9}{4}left(2xright)5(-a)4
Result
Required term T5=4032a4x5
Formula
(iv) General term Tr+1=binom{n}{r}an-rbr,quad n=11, a=x, b=-sqrt{y}
Working
Tr+1=binom{11}{r}x11-r(-sqrt{y})r
Set exponent of y to 3 y3=yr/2 ⇒ r2=3 ⇒ r=6
Result
Required term T6+1=binom{11}{6}x5(-y1/2)6=462x5y3
6.(i) Find the coefficient of x^5 in the expansion of (x^2 - 3/(2x))^10. (ii) Find the coefficient of x^n in the expansion of (x^2 - 1/x)^(2n).
Formula
(i) General term Tr+1=binom{n}{r}an-rbr,quad n=10, a=x2, b=-32x
Working
Tr+1=binom{10}{r}(x2)10-rleft(-32xright)r=binom{10}{r}left(-32right)rx20-3r
Set exponent of x to 5 20-3r=5 ⇒ 15=3r ⇒ r=5
T5+1=binom{10}{5}left(-32right)5x5 ⇒ T6=-153098x5
Result
Coefficient -153098=-1913.625
Formula
(ii) General term Tr+1=binom{n'}{r}an'-rbr,quad n'=2n, a=x2, b=-1x
Working
Tr+1=binom{2n}{r}(x2)2n-rleft(-1xright)r=binom{2n}{r}(-1)r x4n-3r
Set exponent of x to n 4n-3r=n ⇒ 3n=3r ⇒ r=n
Tn+1=binom{2n}{n}(-1)nxn=(2n)!n! n!(-1)nxn
Result
Coefficient (-1)n(2n)!(n!)2
7.Find the 6th term in the expansion of (x^2 - 3/(2x))^10.
Formula
General term Tr+1=binom{n}{r}an-rbr,quad n=10, a=x2, b=-32x
Working
For the 6th term, r=5 T5+1=binom{10}{5}(x2)10-5left(-32xright)5
=252x10 × left(-24332right) × 1x5
Result
6th term T6=-153098x5=-1913.625x5
8.(i) Find the term independent of x in the expansion of (x - 2/x)^10. (ii) Find the term independent of x in the expansion of (sqrt(x) + 1/(2x^2))^10. (iii) Find the term independent of x in the expansion of (1 + x^2)^3 (1 + 1/x^2)^4.
Formula
(i) General term Tr+1=binom{n}{r}an-rbr,quad n=10, a=x, b=-2x
Working
Tr+1=binom{10}{r}x10-rleft(-2xright)r=binom{10}{r}x10-2r(-2)r
Set exponent of x to 0 10-2r=0 ⇒ r=5
T5+1=binom{10}{5}x0(-2)5=10 × 9 × 8 × 7 × 65 × 4 × 3 × 2 × 1(-32)
Result
Required term T6=-8064
Formula
(ii) General term Tr+1=binom{n}{r}an-rbr,quad n=10, a=sqrt{x}, b=12x2
Working
Tr+1=binom{10}{r}(sqrt{x})10-rleft(12x2right)r=binom{10}{r}left(12right)r x5-5r/2
Set exponent of x to 0 5-5r2=0 ⇒ 5r2=5 ⇒ r=2
T3=binom{10}{2}left(12right)2x0=10 × 92 × 14
Result
Required term T3=454
(iii) Combine (1+x2)3left(1+1x2right)4=(1+x2)3(x2+1)4x8=x-8(1+x2)7
Formula
General term Tr+1=x-8binom{7}{r}(1)7-r(x2)r=binom{7}{r}x2r-8
Set exponent of x to 0 2r-8=0 ⇒ r=4
Result
Required term T4+1=binom{7}{4}x0=35
9.(i) Determine the middle term in the expansion of (1/x - x^2/2)^12. (ii) Determine the middle terms in the expansion of (3x/2 - 1/(3x))^11. (iii) Determine the middle terms in the expansion of (2x - 1/(2x))^(2m+1).
Given
(i) Middle term index n=12 (even) ⇒ middle term=left(n2+1right)th=7th term, r=6
Formula
General term Tr+1=binom{n}{r}an-rbr,quad a=1x, b=-x22
Working
T6+1=binom{12}{6}left(1xright)6left(-x22right)6=924left(1x6right)left(frac{x12{64}right)
Result
Middle term T7=23116x6
Given
(ii) Middle term indices n=11 (odd) ⇒ middle terms areleft(n+12right)th=6th and left(n+32right)th=7th terms
Formula
General term Tr+1=binom{11}{r}a11-rbr,quad a=3x2, b=-13x
Working
6th term, r=5 T6=binom{11}{5}left(3x2right)6left(-13xright)5
Result
6th term T6=-69332x
Working
7th term, r=6 T7=binom{11}{6}left(3x2right)5left(-13xright)6=462left(35x525right)left(136x6right)=462left(132 × 3right)1x
Result
7th term T7=7716x
Given
(iii) Middle term indices 2m+1 (odd) ⇒ middle terms are (m+1)th and (m+2)th terms,quad a=2x, b=-12x, n=2m+1
Working
(m+1)th term, r=m Tm+1=binom{2m+1}{m}(2x)m+1left(-12xright)m
Result
(m+1)th term Tm+1=(2m+1)!m!(m+1)!(-1)m 2x
Working
(m+2)th term, r=m+1 Tm+2=binom{2m+1}{m+1}(2x)mleft(-12xright)m+1
Result
(m+2)th term Tm+2=(2m+1)!m!(m+1)!(-1)m+1 12x
10.Show that: C(n,1) + C(n,2) + C(n,3) + ... + C(n,n) = 2^n - 1
Formula
Binomial theorem (1+x)n=binom{n}{0}+binom{n}{1}x+binom{n}{2}x2+cdots+binom{n}{n}xn
Working
Put x = 1 (1+1)n=binom{n}{0}+binom{n}{1}(1)+binom{n}{2}(1)2+cdots+binom{n}{n}(1)n
binom{n}{0}+binom{n}{1}+binom{n}{2}+cdots+binom{n}{n}=2n
Since C(n,0)=1 1+binom{n}{1}+binom{n}{2}+cdots+binom{n}{n}=2n
Result
Hence binom{n}{1}+binom{n}{2}+binom{n}{3}+cdots+binom{n}{n}=2n-1