Mathematics Dictionary

S(n) = (1 + 3 + 5 + .... + n) - (2 + 4 + 6 + ... + 2*(n-1))
First sery : It has m = n/2 terms
- Common difference d = 2
- T(1) = 1
- T(m) = 1 + (m - 1)*d = 1 + (n/2 - 1)*2 = 1 + n - 2 = (n - 1)
- S(m) = m*(T(1) + T(m))/2 = (n/2)*(1 + (n-1))/2 = (n^2)/4
second sery : It has m = n/2 terms
- Common difference d = 2
- T(1) = 2
- T(m) = 2 + (m - 1)*d = 2 + (n/2 - 1)*2 = 2 + n - 2 = n
- S(m) = m*(T(1) + T(m))/2 = (n/2)*(2 + n))/2 = (n*(n+2))/4
Hence S(n) = (n^2)/4 - (n^2 + 2*n)/4 = -n/2

Second Method : Conjecture

n = 2 : S(2) = 1 - 2 = -1
n = 4 : S(4) = 1 - 2 + 3 - 4 = -2
n = 6 : S(6) = 1 - 2 + 3 - 4 + 5 - 6 = -3
Hence S(n) = -n/2

Exercise

Find S(n) if n is odd (See AL 08 12)

Go to Begin

AL 08 08. Sum[(2*n-1)^2] = n*(4*n^2 - 1)/3
Induction method

n = 1, LHS = 1 and RHS = 1
n = 2, RHS = 10 and RHS = 2*(16 - 1)/3 = 10
....
Both sides add (2*n + 1)
We expect RHS = ((n+1)*(4*(n+1)^2 - 1)/3
Proof
- RHS = n*(4*n^2 - 1)/3 + (2*n + 1)^2
- RHS = (4*n^3 - n) + (12*n^2 +12*n + 3))/3
- RHS = (4*n^3 + 12*n^2 + 11*n + 3))/3
- RHS = ((n+1)*(4*N^2 + 8*n +3))/3
- RHS = (n+1)*(4*(n+1)^2 - 1)/3

Other method

Sum[(2*n-1)^2] = Sum[(4*n^2 - 4*n + 1)]
Use formula
- Sum[1] = n
- Sum[n] = n*(n+1)/2
- Sum[n^2] = n*(n+1)*(2*n+1)/6
Sum[(2*n-1)^2] = Sum[(4*n^2 - 4*n + 1)]
= Sum[4*n^2] - Sum[4*n] + Sum[1]
= 4*n*(n+1)*(2*n+1)/6 - 4*n*(n+1)/2 + n
= n*(4*(n+1)*(2*n+1)/6 - 2*n - 2 + 1)
= n*(8*n^2 + 12*n + 4 - 12*n - 12 + 6)/6
= n*(8*n^2 - 2)/6
= n*(4*n^2 - 1)/3

Go to Begin

AL 08 09. Sum[n*((n+1)^2)] = n*(n+1)*(n+2)*(3*n+5)/12

Induction

n = 1, LHS = 1*2^2 = 4 and RHS = (1*2*3*8)/12 = 4
n = 2, LHS = 4 + 2*9 = 22 and RHS = (2*3*4*11)/12 = 22
...
Both sides add (n+1)*(n+2)^2
We expect (n+1)*((n+1)+1)*((n+1)+2)*(3*(n+1)+5)/12
Proof

Other method

Sum[n*(n+1)^2] = Sum[(n^3 + 2*n^2 + n)]
Use formula
- Sum[n] = n*(n+1)/2
- Sum[n^2] = n*(n+1)*(2*n+1)/6
- Sum[n^3] = (n*(n+1)/2)^2
Sum[n^3 + 2*n^2 +n]
= Sum[n^3] + Sum[2*n^2] + Sum[n]
= (n*(n+1)/2)^2 + 2*n(n+1)*(2*n+1)/6 + n*(n+1)/2
= n*(n+1)*(n*(n+1)/4) + (2*n+1)/3 + 1/2
= n*(n+1)*(3*n^2 + 3*n + 8*n + 4 + 6)/12
= n*(n+1)*(3*n^2 - 11*n + 10)/12
= n*(n+1)*(n+2)*(3*n+5)/12

Go to Begin

AL 08 10. 1^3 + 2^3 + ... + n^3 GT (n^4)/4 GT 1^3 + 2^3 + ... +(n-1)^3

Keywords

Sum[n^3] = ((n*(n+1)/2)^2

Proof

Sum[n^3] = ((n^2 + n)^2)/4 = (n^4 + 2*n^3 + n^2)/4
Hence Sum[n^3] GT (n^4)/4
Sum[(n-1)^3] = ((n-1)*n)/2)^2 = (n^2 - n)^2/4 = (n^4 - n*(2*n - 1))/4
Hence n^4 GT Sum[(n-1)^3]

Go to Begin

AL 08 11. S(n)=(1-1/4)*(1-1/9)*(1-1/16)*....*(1-1/((n+1)^2 = (n+1)/2*n. n GT 1

Conjective method

S(2) = 1-1/4 = 3/4
S(3) = (3/4)*(1-1/9) = 4/6
S(4) = (4/6)*(1-1/16) = (4/6)*(15/16) = 5/8
S(n) = (n+1)/(2*n). Need to prove this by induction

Go to Begin

AL 08 12. S(n) = 1 - 2 + 3 - 4 + ...... + n and n is odd

Keywords

Use AP formula

Solution

S(n) = (1 + 3 + 5 + .... + (2*n-1)) - (2 + 4 + 6 + ... + 2*(n-1))
First sery : It has m = (n+1)/2) terms
- Common difference d = 2
- T(1) = 1
- T(m) = 1 + (m - 1)*d
- = 1 + ((n+1)/2 - 1)*2
- = 1 + n + 1 - 2
- = n
- S(m) = m*(T(1) + T(m))/2
- = ((n+1)/2)*(1 + n))/2
- = ((n+1)^2)/4
- = (n^2 + 2*n +1)/4
second sery : It has m = (n-1)/2 terms
- Common difference d = 2
- T(1) = 2
- T(m) = 2 + (m - 1)*d
- = 2 + ((n-1)/2 - 1)*2
- = 2 + n - 1 - 2
- = n - 1
- S(m) = m*(T(1) + T(m))/2
- = ((n-1)/2)*(2 + (n-1))/2
- = ((n-1)^2)/4 + 2*(n-1)/4
- = (n*2 - 2*n + 1 + 2*n - 2)/4
- = (n^2 - 1)/4
Hence S(n) = (n^2 + 2*n + 1)/4 - (n^2 - 1)/4 = (n+1)/2

Second Method : Conjecture

n = 3 : S(3) = 1 - 2 + 3 = 2
n = 5 : S(5) = 1 - 2 + 3 - 4 + 5 = 3
n = 7 : S(7) = 1 - 2 + 3 - 4 + 5 - 6 + 7 = 4
Hence S(n) = (n+1)/2

Exercise

Find S(n) if n is even (See AL 08 07)

Go to Begin

AL 08 13. pi in series

1. pi/4 = 1 - 1/3 + 1/5 + 1/7 - ...
2. ((pi)^2) = 6*(1 + 1/(2^2) + 1/(3^2) + ...)
3. ((pi)^2) = 8*(1 + 1/(3^2) + 1/(5^2) + ...)

Reference

Topic | Story of pi

Go to Begin

AL 08 14. AP : Angles of triangle ABC are 3 consecutive AP terms. Find angles

Solution

Let angle B = x, then angle A = x - d and angle C = x + d where d is common diff
Hence Angles A + B + C = 3*x = 180 degrees
Hence x = 60 degress and angle B = 60
Find d
- A = 60 - d
- C = 60 + d
- A + C = 120
- Hence d = 30 by trial
Nence angle A = 30 and C = 90 degrees

Go to Begin

AL 08 15. GP : Angles of triangle ABC are 3 consecutive GP terms. Find angles

Solution

Let angle A = x, then angle B = r*x and angle C = x*(r^2) where r is common ratio
Hence Angles A + B + C = x*(1 + r + r^2) = 180 degrees
Hence r^2 + r + 1 = 180/x
Hence r^2 + r + (1 - 180/x) = 0
Solve for r
- r = (-1 + Sqr(1 - 4*(1 - 180)/x))/2 or r = (-1 - Sqr(1 - 4*(1 - 180)/x))/2
- If r is given, then we can find A,B,C
- If r is not given
  - 1 - 4*(1 - 180/x) must be positive
  - Hence r can be positive or negative real numbes

Example : Find angles if r = 3

3 = (-1 + Sqr(1 - 4*(1 - 180/x))/2
6 + 1 = Sqr(1 - 4*(1 - 180/x)
Square both sides
- 49 = 1 - 4*(1 - 180/x)
- 48 = -4*(1 - 180/x)
- 12 = -1 + 180/x
- 13*x = 180
- Hence A = x = 13.846, B = x*r = 41.538 and C = x*(r^2) = 124.615
Verify : A + B + C = 179.999 degrees

Go to Begin

AL 08 16. Compound interest

Formula

Let P = principal, i = interest, n = number of interest period
The formula is S = P*(1 + i)^n

Interest periods

Compound Monthly with annual rate i for n years : S = P*(1 + i/12)^(12*n)
Compound semi-annually with annual rate i for n years : S = P*(1 + i/2)^(2*n)
Compoun annualy with annual rate i for n years : S = P*(1 + i)^(n)

Proof of the formula

1st period : S1 = P*(1 + i)
2nd period : S2 = S1 + i*S1 = S1*(1 + i) = P*(1 + i)^2
3rd period : S3 = S2 + i*S2 = S2*(1 + i) = P*(1 + i)^3
4th period : S4 = S3 + i*S3 = S3*(1 + i) = P*(1 + i)^4
nth period : S = P*(1 + i)^n

Question : How to calculate 1.2^10 ?

If calculator has function y^x
- Give number 1.2
- Press y^x
- Give number 10
- Press equal sign (=) : 1.2^10 = 6.1917
If calculator has no function y^x, we can use ln(x)
- Let a = 1.2^10
- Take log on both sides : ln(a) = 10*ln(1.2) = 1.8232
- Hence a = e^(1.8232) = 6.1916

Go to Begin

AL 08 17. Deposit fixed amount with fixed rate and fixed period

Formulate a formula