Q01 |
- Sequences : 2 6 18 54 .... find 10th term
Q02 |
- Sequences : 1, 3, 6, 10, ..... find nth term
Q03 |
- Properties of numbers sequence in triangular pattern
Q04 |
- Prove that Sum[n*(n+1)/2] = n*(n+1)*(n+2)/3! = C(n+2,3)
Q05 |
- Numbers matrix pattern : First row numbers are in triangle number sequences
Q06 |
- Pattern in Q5, find the number at row r and column c
Q07 |
- Use Q5 pattern finding row and column for number 11
Q08 |
- Find more sequences
Q09 |
- Pascal triangle and sequences in MD
Q10 |
- Formula
Q11 |
- Reference for sequences in Pascal triangle
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Answers
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Q01. Sequences : 2 6 18 54 .... find 10th term
Solution
* 2*( 1, 3, 9, 27, ....)
* 2*( 3^0, 3^1, 3^2, 3^3, ....)
* 10th term = 2*(3^9) = ?
Find Sum[n^3]
- This is the 3 to power n sequence (3^n) for n starting with 0
- Find the sum of the nth term
- More about Sequence 1^3, 2^3, 3^3, .... n^3
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Q02. Sequence : 1, 3, 6, 10, ..... find nth term
Solution
This is the sequence of numbers in triangular sequence
* .................. 1 = 1*(1+1)/2
*
* * ................ 3 = 2*(2+1)/2
*
* *
* * * .............. 6 = 3*(3+1)/2
*
* *
* * *
* * * * ........... 10 = 4*(4+1)/2
Hence : T(n) = n*(n+1)/2!
Hence : S(n) = n*(n+1)*(n+2)/3!
Prove that Sum[n*(n+1)/2] = n*(n+1)*(n+2)/3!
- Use Sum[n] = n*(n+1)/2
- Use Sum[n*2] = n*(n+1)*(2*n+1)/6
- Proof :
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Q03. Properties of numbers sequence in triangular pattern
* 1st difference F(k) = T(k+1)-T(K) : 2 3 4 5 6 ......
* 2nd difference G(k) = F(k+1)-F(k) : 1 1 1 1 1 ......
* T(n) = n*(n+1)/2 = C(n+1,2)
* S(n) = Sum[n*(n+1)/2] = n*(n+1)*(n+2)/6 (Proof : See Q4)
* S(n) = n*(n+1)*(n+2)/3! = C(n+2,3) : Proof
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* This is same as quadratic function y=x*(x+1)/2
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Q04. Prove that Sum[n*(n+1)/2] = n*(n+1)*(n+2)/3! = C(n+2,3)
Proof
1. Sum[n*(n+1)/2] = Sum[n*2]/2 + Sum[n]/2
2. Sum[n^2] = n*(n+1)*(2*n+1)/6
3. Sum[n^1] = n*(n+1)/2
4. Substitute 2 and 3 into 1 we have
4. Sum[n*(n+1)/2] = n*(n+1)*(2*n+1)/12 + n(n+1)/4
5. Sum[n*(n+1)/2] = n*(n+1)*((2*n+1)/12 + 1/4)
6. Sum[n*(n+1)/2] = n*(n+1)*(n+2)/6 = C(n+2,3)
- This sequences can be found in Pascal Triangle on internet (See Q11)
- Where C(n,r) is the coefficient of rth term in expansion of (a+b)^n
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Q05. Numbers matrix pattern : First row numbers are in triangle number sequences
Questions
- 1. Find number if 10th row and 10 column
- 2. Find the coulmn and row number for number 100
Solution
View Pattern and application.
(1) Find number at given row and given column.
(2) Find row and column for a given number.
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Q06. Pattern in Q5, find the number at row r and column c
Formula and exercise
* Formula : Number = 1 + 2 + 3 + .... (r+c-2) + c
* Example : Find number at row 1 and colum 4.
- Row r = 1 and column c = 4
- (r+c-2) = 3 Hence 1 + 2 + 3 + 4 = 10
* Exercise : Find number at row 6 and column 9
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Q07. Use Q5 pattern finding row and column for number 11
Solution
* Numbers at Row 1 : 1, 3, 6, 10, 15, 21, .....
* At row 1 and column x the number is x*(x+1)/2
* Hence x*(x+1)/2 must be greater than 11
* Hence x = 4, x*(x+1)/2 = 10
* Hence x = 5, x*(x+1)/2 = 15. This is the required x
* Number at row 1 and column x is 15
* Hence row number for 11 is r = (15 - 11) = 4
* Hence col number for 11 is c = x - (15-11) = 1
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Q08. Find more sequences
Sequences
Keyword in MD : Fibonacci's sequence
Keyword in MD : Keppler's defintion for Fibonacci's sequence
Keyword in MD : Lucas sequences
Keyword in MD : Sequences
Keyword in MD : Pascal triangle and sequences
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Q09. Pascal triangle and sequence in MD
Examples
Example 1 : Sum[C(n+1, 2)] = C(n+2, 3)
Example 2 : Sum[C(n+2, 3)] = C(n+3, 4)
Example 3 : Sum[C(n+3, 4)] = C(n+4, 5)
Example 4 : Sum[C(n+4, 5)] = C(n+5, 6)
Proof can be found on internet
Click here
Pascal Triangle
r = ......0 ... 1 ... 2 ... 3 ... 4 ... 5 ...
n = 0 ... 1
n = 1 ... 1 ... 1
n = 2 ... 1 ... 2 ... 1
n = 3 ... 1 ... 3 ... 3 ... 1
n = 4 ... 1 ... 4 ... 6 ... 4 ... 1
n = 5 ... 1 ... 5 .. 10 .. 10 ... 5 ... 1
n = 6 ... 1 ... 6 .. 15 .. 20 .. 15 ... 6 ... 1
n = 7 ... 1 ... 7 .. 21 .. 35 .. 35 .. 21 ... 7 ... 1
Applications
Note 1 : when r=2, it is triangular number sequence in Example 1
Note 2 : when r=3, sequence 1, 4, 10, 20, .... in Example 2
Note 3 : when r=4, sequence 1, 5, 15, 35, .... in Example 3
Illustration
- Along line r=2, numbers 1, 3, 6, 10, 15
- Sum the number = 1 + 3 + 6 + 10 + 15 = 35
- We can find 35 along line r=3 which is at C(7,3) i.e. r=3 and n=7
- Hence sum of number along line at r=2 can be found at along line r=3
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Q10. Formula
- Sum[1] = n if there are n terms
- 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*(n+1)/2] = n*(n+1)*(n+2)/3!
- Arithmetic Progression (A. P.)
- T(1) = a is the first term
- T(n)=a+(n-1)*d where d is common difference = T(n)-T(n-1)
- S(n) = n*(T(1)+T(n))/2
- a,b,c in arithmetric sequence then b-a = c-b and b is arithmetric mean
- Geometirc progression
- T(1) = a
- T(n) = a*r^(n-1) where r is common ration = T(n)/T(n+1)
- S(n) = a*(1-r^n)/(1-r)
- If r < 1 then S(n) = a/(1-r) and the series is convergent
- if r >= 1 then sequence is diverge
- a,b,c in geometric sequence the a/b = b/c and b is geometric mean
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Q11. Reference
Pascal triangle and sequences
- On
Internet
Pascal Triangle and sequences
- Keyword in MD : Sequence and Pascal triangle
- Keyword in MD : Series and sequences
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