Read Symbol defintion
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Q01 |
- Highlights
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Q02 |
- Method 1 : Prove that Sum[n^2)] = n*(n+1)*(2*n+1)/6
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Q03 |
- Method 2 : Prove that Sum[n^2)] = n*(n+1)*(2*n+1)/6
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Q04 |
- Method 3 : Prove that Sum[n^2)] = n*(n+1)*(2*n+1)/6
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Q05 |
- Method 4 : Prove that Sum[n^2)] = n*(n+1)*(2*n+1)/6
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Q06 |
- Patterns : Squres in squares
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Q07 |
- Pattern : Numbers in square patterns
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Q08 |
- Second difference of squaence 1, 4, 9, 16, 25, .....
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Q09 |
- Home work
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Q10 |
- Exercises
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Q11 |
- Exercises
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Answers
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Q01. Highlight
- Pattern : Squares in squares
- Pattern : Numbers in square patterns
- Proof of the formula using 4 different methods
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Q2 Prove that 1^2 + 2^2 + 3^2 +...+ n^2 = n*(n+1)*(2*n+1)/6
* Method 1 : By observation
1^2 + 2^2 = 1 + 4 = 5 = 2*3*5/6
1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14 = 3*4*7/6
1^2 + 2^2 + 3^2 + 4^2 = 1 + 4 + 9 + 16 = 30 = 4*5*9/6
Since 4*5*9/6 = 4*(4+1)*(2*4+1)/6 for n = 4
Proof complete
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Q03. Prove that 1^2 + 2^2 + 3^2 +...+ n^2 = n*(n+1)*(2*n+1)/6
Method 2 : Use Sum[C(n+1,2)] = C(n+3,2)
C(n+1,2) = (n+1)*n/2!
C(n+2,3) = (n+2)*(n+1)*n/3!
Sum[(n+1)*n/2!] = (n+2)*(n+1)/3!
Expnand :
Sum[n^2 + n] = 2*(n+2)*(n+1)*n/6
Hence Sum[n^2] = 2*(n+2)*(n+1)*n/6 -Sum[n]
Since Sum[n] = n*(n+1)/2
Hence Sum[n^2] = n*(n+1)*[2*(n+2)/6 - 1/2]
Hence Sum[n^2] = n*(n+1)*(2*n+1)/6
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Q04. Prove that 1^2 + 2^2 + 3^2 +...+ n^2 = n*(n+1)*(2*n+1)/6
Method 3 : Use mathematical induction
n = 1 the sum is true
n = 2 the sum is true
n = k+1 should be true. i.e. S(k+1)=(k+1)*(k+2)*(2*(k+1)+1)/6 exists
n = k+1 and S(k+1) = S(k) + (k+1)^2
Hence S(k+1) = k*(k+1)*(2*k+1)/6 + (k+1)^2
Hence S(k+1) = (k+1)*(k*(2*k+1)/6 + (k+1))
Hence S(k+1) = ((k+1)*(2*k^2+k)/6 + k+1))
Hence S(k+1) = (k+1)*(2*k^2 + 7*k + 6)/6
Hence S(k+1) = (k+1)*(k+1+1)*(2*(k+1)+1)/6
Proof complete
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Q05. Prove that 1^2 + 2^2 + 3^2 +...+ n^2 = n*(n+1)*(2*n+1)/6
Method 4 : Use (a+b)^3 = a^3 + 3*(a^2)*b + 3*a*(b^2) + 1 ............(1)
- Since (x+1)^3 = x^3 + 3*x^2 + 3*x + 1
- Sum[(x + 1)^3 - x^3] = Sum[3*x^2] + Sum[3*x] + Sum[1]
- Sum[(x + 1)^3 - x^3]
- = (2^3 - 1) + (3^3 - 2^3) + ...... ((x+1)^3 - x^3)
- = (x + 1)^3 - 1 (All terms cancelled out expcept -1 and (x+1)^3)
- = x^3 + 3*x^2 + 3*x
- Hence Sum[3*x^2] + Sum[3*x] + Sum[1] = x^3 + 3*x^2 + 3*x ...... (2)
- From (1) and (2) we have
- Sum[3*x^2] + Sum[3*x] + Sum[1] = x^3 + 3*x^2 + 3*x
- Sum[3*x^2] = x^3 + 3*x^2 + 3*x - Sum[3*x] - Sum[1]
- Sum[3*n^2] = n^3 + 3*n^2 + 3*n - 3*n*(n+1)/2 - n
- = n^3 + 3*n^2 + 3*n - 3*n*(n+1)/2 - n
- = n^3 + 2*n + 3*n^2 - 3*n^2/2 - 3*n/2
- = n^3 + 3*n^2/2 + n/2
- = n*(2*n^2 + 3*n + 1)/2
- = n*(n+1)*(2*n+1)/2
- Hence Sum[n^2] = n*(n+1)*(2*n+1)/6
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Q06. Pattern :
Squares in squares
1 + 4 + 9 + 16 + ....
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Q07. Pattern : Numbers in square patterns
* .......... 1^2 = 1
* *
* * ........ 2^2 = 4
* * *
* * *
* * * ...... 3^2 = 9
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Q08. Find 2nd difference of sequence 1, 4, 9, 16, 25, 36,....
Sequence : 01 04 09 16 25 36 .....
1st diff : .. 03 05 07 09 11 .....
2nd diff : ..... 02 02 02 02 .....
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Q09. Home work :
a. Write all formulae used in above methods to note book.
b. Try to remember them
c. Fid more series which are given in Chapter 14 of MD2002
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Q10. Exercises :
a. Prove Sum[n^1] = (n*(n+1))/2 ........... See program 14 01
b. Prove Sum[n^2] = (n*(n+1)*(2*n+1)/6 .... See Program 14 12
c. Sequence 1, 4, 09, 16, 25, ...... Find the 10th term
c. Sequence 2, 5, 10, 17, 26, ...... Find the 10th term
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Q11. Reference
- MD2002 Program 14 13
- MD2002 text file ZS.txt
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