StudentAidUnit 8: Mathematical Inductions and Binomial Theorem — Exercise 8 3
11th Class Mathematics · Unit 8: Mathematical Inductions and Binomial Theorem
1.(i) Expand the following up to 4 terms, taking the value of x such that the expansion in each case is valid: (1+x)^(-1/3) (ii) Expand the following up to 4 terms: (4-3x)^(1/2) (iii) Expand: (1-x)^(-1) / (1+x)^(-2) (iv) Expand: sqrt(1+2x) / (1-x)
Formula
(i) Binomial expansion
(1+x)n = 1 + nx + n(n-1)2!x2 + n(n-1)(n-2)3!x3 + ...
Working
With n = -1/3
(1+x)-1/3 = 1 + left(-13right)x + (-1/3)(-1/3-1)2!x2 + (-1/3)(-1/3-1)(-1/3-2)3!x3 + ...
= 1 - 13x + (-1/3)(-4/3)2x2 + (-1/3)(-4/3)(-7/3)6x3 + ...
= 1 - 13x + 29x2 - 1481x3 + ...
Result
Expansion
(1+x)-1/3 = 1 - 13x + 29x2 - 1481x3 + ...
Valid when
|x| < 1
(ii) Factor out 4
(4-3x)1/2 = 41/2left(1-3x4right)1/2 = 2left(1-3x4right)1/2
Formula
Apply binomial with n=1/2, x=-3x/4
= 2left[1 + 12left(-3x4right) + (1/2)(-1/2)2!left(-3x4right)2 + (1/2)(-1/2)(-3/2)3!left(-3x4right)3 + ...right]
= 2left[1 - 3x8 + (1/2)(-1/2)2 · 9x216 + (1/2)(-1/2)(-3/2)6 · left(-27x364right) + ...right]
= 2left[1 - 3x8 - 9x264 - 27x3512 + ...right]
Result
Expansion
= 2 - 3x4 - 9x264 - 27x3512 + ...
Valid when
left|3x4right| < 1 ⇒ |x| < 43
(iii) Rewrite
frac{(1-x)-1{(1+x)-2 = (1-x)-1(1+x)2 = [(1-x)-1][(1+x)2]
Formula
Expand (1-x)^(-1)
(1-x)-1 = 1 + (-1)(-x) + (-1)(-1-1)2!(-x)2 + ... = 1 + x + x2 + x3 + ...
Expand (1+x)^2
(1+x)2 = 1 + 2x + x2
Multiply series
[1 + x + x2 + x3 + ...][1 + 2x + x2]
= 1 + x + x2 + x3 + ... + 2x + 2x2 + 2x3 + ... + x2 + x3 + ...
Result
Expansion (up to x³)
= 1 + 3x + 3x2 + 2x3 + ...
Valid when
|x| < 1
(iv) Rewrite
frac{sqrt{1+2x}{1-x} = (1+2x)1/2(1-x)-1
Formula
Expand (1+2x)^(1/2)
(1+2x)1/2 = 1 + 12(2x) + (1/2)(-1/2)2!(2x)2 + ... = 1 + x - x22 + x32 + ...
Expand (1-x)^(-1)
(1-x)-1 = 1 + x + x2 + x3 + ...
Multiply series
left(1 + x - x22 + x32 + ...right)(1 + x + x2 + x3 + ...)
= 1 + x + x2 + x + x2 + x3 - x22 - x32 + ...
Result
Expansion (up to x³)
= 1 + 2x + 3x22 + 2x3 + ...
Valid when
|x| < 12 and |2x| < 1
2.(i) Find the coefficient of x^n in the expansion of (1+x²)/(1+x)² (ii) Find the coefficient of x^n in the expansion of (1+x)²/(1-x)²
(i) Rewrite
1+x2(1+x)2 = (1+x2)(1+x)-2
Formula
General term of (1+x)^n
Tr+1 = n(n-1)(n-2)...(n-r+1)r!xr
For (1+x)^(-2), coefficient of x^n
Tn+1 = (-1)n(n+1)xn
Coefficient of x^(n-2) in (1+x)^(-2)
Tn-1 = (-1)n-2(n-1)xn-2
Term from (1+x²)·(1+x)^(-2)
Coefficient of xn = (-1)n(n+1) + (-1)n-2(n-1)
Result
Coefficient of x^n
= (-1)n(n+1) + (-1)n-2(n-1) = (-1)n(2n)
(ii) Rewrite
(1+x)2(1-x)2 = (1+2x+x2)(1-x)-2
Formula
General term of (1-x)^(-2)
Tr+1 = (r+1)xr
Coefficient of x^n from 1·(1-x)^(-2)
(n+1)xn
Coefficient of x^(n-1) from 2x·(1-x)^(-2)
2 · n · xn
Coefficient of x^(n-2) from x²·(1-x)^(-2)
(n-1)xn
Sum of all contributions
= (n+1) + 2n + (n-1) = 4n
Result
Coefficient of x^n
4n
3.(i) If x is so small that its square and higher powers can be neglected then show that (1-x)/sqrt(1+x) ≈ 1 - (3/2)x (ii) If x is so small that its square and higher powers can be neglected then show that sqrt(1+2x)/sqrt(1-x) ≈ 1 + (3/2)x (iii) If x is so small that its square and higher powers can be neglected then show that [(9+7x)^(1/2) - (16+3x)^(1/4)] / (4+5x) ≈ (1/4) - (17/384)x (iv) If x is so small that its cube and higher power can be neglected, then show that sqrt(4+x)/(1-x)³ ≈ 2 + (25/4)x
Given
(i) LHS
frac{1-x}{sqrt{1+x} = (1-x)(1+x)-1/2
Formula
Expand (1+x)^(-1/2)
(1+x)-1/2 = 1 + left(-12right)x + ...
Neglecting x²
≈ 1 - 12x
Multiply by (1-x)
(1-x)left(1 - 12xright) ≈ 1 - 12x - x + 12x2
Neglecting x²
≈ 1 - 32x
Result
Result
frac{1-x}{sqrt{1+x} ≈ 1 - 32x
Given
(ii) LHS
frac{sqrt{1+2x}{sqrt{1-x} = (1+2x)1/2(1-x)-1/2
Formula
Expand (1+2x)^(1/2)
(1+2x)1/2 ≈ 1 + 12(2x) = 1 + x
Expand (1-x)^(-1/2)
(1-x)-1/2 ≈ 1 + 12x
Multiply (1+x)(1+x/2)
≈ (1+x)left(1 + x2right)
Neglecting x²
≈ 1 + x + x2 = 1 + 32x
Result
Result
frac{sqrt{1+2x}{sqrt{1-x} ≈ 1 + 32x
Given
(iii) Numerator expansion
(9+7x)1/2 = 91/2left(1+7x9right)1/2 ≈ 3left(1+7x18right) = 3 + 7x6
(16+3x)1/4 = 161/4left(1+3x16right)1/4 ≈ 2left(1+3x64right) = 2 + 3x32
Difference
left(3 + 7x6right) - left(2 + 3x32right) = 1 + 7x6 - 3x32 = 1 + 103x96
Denominator expansion
14+5x = 14left(1-5x4right) ≈ 14left(1-5x4right)
Product to first order in x
left(1 + 103x96right) · 14left(1-5x4right) ≈ 14left(1 - 5x4 + 103x96right)
= 14 - 5x16 + 103x384 = 14 + -120x + 103x384 = 14 - 17x384
Result
Result
≈ 14 - 17x384
Given
(iv) LHS
frac{sqrt{4+x}{(1-x)3} = (4+x)1/2(1-x)-3
Factor out 4 from square root
= 41/2left(1+x4right)1/2(1-x)-3
Formula
Expand (1+x/4)^(1/2)
≈ 2left(1+12 · x4right) = 2left(1+x8right)
Expand (1-x)^(-3)
≈ 1 + 3x
Multiply
2left(1+x8right)(1+3x) ≈ 2left(1 + x8 + 3xright) = 2left(1 + 25x8right)
Result
Result
≈ 2 + 25x4
4.(i) If x is so small that its cube and higher power can be neglected, then show that sqrt(1-x-2x²) ≈ 1 - (1/2)x - (9/8)x² (ii) If x is so small that its cube and higher power can be neglected, then show that sqrt((1+x)/(1-x)) ≈ 1 + x + (1/2)x²
Given
(i) LHS
sqrt{1-x-2x2} = (1-(x+2x2))1/2
Formula
Binomial expansion
= 1 + 12(-(x+2x2)) + (1/2)(-1/2)2!(-(x+2x2))2 + ...
Neglecting x³
= 1 - 12(x+2x2) + frac{-1/8}(x2+...) + ...
= 1 - 12x - x2 - 18x2 + ...
= 1 - 12x - 98x2 + ...
Result
Result
sqrt{1-x-2x2} ≈ 1 - 12x - 98x2
Given
(ii) LHS
sqrt{1+x1-x = sqrt{(1+x)(1-x)-1 = (1+x)1/2(1-x)-1/2
Formula
Expand (1+x)^(1/2)
≈ 1 + 12x
Expand (1-x)^(-1/2)
≈ 1 + 12x
Multiply
left(1 + 12xright)left(1 + 12xright) = 1 + 12x + 12x + 14x2
Simplify
= 1 + x + 14x2
Result
Result (up to x²)
sqrt{1+x1-x ≈ 1 + x + 12x2
5.If x is nearly equal to 1, then prove that px^p - qx^q ≈ (p-q)x^(p+q)
Given
Let x = 1+h where h is small
As x is nearly equal to 1, let x=1+h where h2, h3, ... are neglected
LHS
L.H.S = p(1+h)p - q(1+h)q
Formula
Expand (1+h)^p
= p[1 + ph + ...] - q[1 + qh + ...]
Neglecting h², h³,...
≈ p + p2h - q - q2h
Simplify
= (p-q) + (p2-q2)h
= (p-q) + (p-q)(p+q)h
= (p-q)[1 + (p+q)h]
RHS as x^(p+q)
= (p-q)(1+h)p+q
Result
Result
Hence, pxp - qxq ≈ (p-q)xp+q
6.(i) Identify the following series as binomial expansion and find the sum: 1 - (1/2)(1/4) + (1.3)/(2!·4)(1/4)² - (1.3·5)/(3!·8)(1/4)³ + ... (ii) Use binomial theorem to show that 1 + 1/4 + 1.3/(4.8) + 1.3·5/(4.8·12) + ... = √2
Given
(i) Let (1+x)^n equal the series
Let (1+x)n = 1 - 12left(14right) + 1.32! · 4left(14right)2 - 1.3 · 53! · 8left(14right)3 + ...
Formula
Binomial expansion formula
(1+x)n = 1 + nx + n(n-1)2!x2 + n(n-1)(n-2)3!x3 + ...
Comparing nx with -1/2·(1/4)
nx = -12 · 14 ⇒ nx = -18
Comparing second term
n(n-1)x2/2! = 1.3/(2!·4)·(1/4)2
From these conditions
nleft(14right) = -18 and n(n-1)2left(116right) = 1.38 · 16
Solving
x = 14, quad n = -12
The series represents
left(1+14right)-1/2 = left(54right)-1/2 = left(45right)1/2 = frac{2}{sqrt{5}
Result
Sum of series
= frac{2}{sqrt{5} or frac{2sqrt{5}{5}
Given
(ii) Let (1+x)^n = series
Let (1+x)n = 1 + 14 + 1.34.8 + 1.3 · 54.8 · 12 + ...
Formula
General term of binomial expansion
(1+x)n = 1 + nx + n(n-1)2!x2 + n(n-1)(n-2)3!x3 + ...
Comparing first coefficient
nx = 14
Comparing second coefficient
n(n-1)2x2 = 1.34.8
Solving the system
nleft(-12right) = 14 and x = -12 ⇒ n = -12
left(1-12right)-1/2 = left(12right)-1/2 = 21/2 = sqrt{2}
Result
Sum equals
sqrt{2}
7.(i) If y = 1/3 + (1.3)/(2!)·(1/3)² + (1.3·5)/(3!)·(1/3)³ + ... Then prove that y² + 2y - 2 = 0 (ii) If 2y = 1/2² + (1.3)/(2!2⁴) + (1.3·5)/(3!2⁶) + ... Then prove that 4y² + 4y - 1 = 0
Given
(i) Given series for y
y = 13 + 1.32!left(13right)2 + 1.3 · 53!left(13right)3 + ...
Add 1 to both sides
1 + y = 1 + 13 + 1.32!left(13right)2 + 1.3 · 53!left(13right)3 + ...
Formula
This is binomial series (1+x)^n
Let (1+x)n = 1 + 13 + 1.32!left(13right)2 + ...
Comparing terms
nx = 13, quad n(n-1)2x2 = 1.32!left(13right)2
Solving
nleft(-23right) = 13 ⇒ n = -12, quad x = -23
Therefore
1 + y = left(1-23right)-1/2 = left(13right)-1/2 = 31/2
Squaring both sides
(1+y)2 = 3
Expanding
1 + 2y + y2 = 3
Result
Required result
y2 + 2y - 2 = 0
Given
(ii) Given series for 2y
2y = 122 + 1.32!24 + 1.3 · 53!26 + ...
Add 1 to both sides
1 + 2y = 1 + 122 + 1.32!24 + 1.3 · 53!26 + ...
Formula
This is binomial series
Let (1+x)n = 1 + 122 + 1.32!24 + ...
Comparing
nleft(-12right) = 14 ⇒ n = -12, quad x = -12
Therefore
1 + 2y = left(1-12right)-1/2 = left(12right)-1/2 = sqrt{2}
So (1+2y)² = 2
(1+2y)2 = 2
Expanding
1 + 4y + 4y2 = 2
Result
Required result
4y2 + 4y - 1 = 0
8.Show that the coefficient of x^r in x/[(1-px)(1-qx)] is (p^r - q^r)/(p-q)
Given
Partial fraction decomposition
x(1-px)(1-qx) = A1-px + B1-qx
Multiply by (1-px)(1-qx)
x = A(1-qx) + B(1-px)
Put 1-px=0, x=1/p
1p = Aleft(1-qpright) ⇒ A = 1p-q
Put 1-qx=0, x=1/q
1q = Bleft(1-pqright) ⇒ B = -1p-q
So
x(1-px)(1-qx) = 1p-qleft(11-px - 11-qxright)
Expand as series
= 1p-qleft[(1+(px)+(px)2+...)-(1+(qx)+(qx)2+...)right]
= 1p-qleft[(p-q)x + (p2-q2)x2 + ... + (pr-qr)xr + ...right]
Result
Coefficient of x^r
= pr - qrp-q