Highlight
- Sum[n^1] = n*(n+1)/2
- Sum[n^2] = n*(n+1)*(2*n+1)/6
- Sum[n^3] = (n*(n+1)/2)^2
Q1. Deifintion of Pascal Triangle
A1. Answer
...... r0 r1 r2 r3 r4 r5 r6 r7 r8
---------------------------------
n0 ... 01
n1 ... 01 01
n2 ... 01 02 01
n3 ... 01 03 03 01
n4 ... 01 04 06 04 01
n5 ... 01 05 10 10 05 01
n6 ... 01 06 15 20 15 06 01
n7 ... 01 07 21 35 35 21 07 01
n8 ... 01 08 28 56 70 56 28 08 01
Q2. What is r1 series ?
A2. Answer
- r1 series : 1, 2, 3, 4, 5, 6, 7, 8, ....
- T(n) = nth term = n
- S(n) = Sum[n^1] = n*(n+1)/2
- 1st difference is common difference d=1
Q3. What is r2 series ?
A3. Answer
- r2 series : 1, 3, 6, 10, 15, 21, 28, 36, ....
- T(n) = nth term = n*(n+1)/2 = C(n+1, 2)
- S(n) = Sum[n^1] = n*(n+1)*(n+2)/6 = C(n+2, 3)
- 1st difference is r1 series
- 2nd difference is common difference d=1
Q4. What is r3 series ?
A4. Answer
- r3 series : 1, 4, 10, 20, 35, 56, ....
- T(n) = nth term = n*(n+1)*(n+2)/3! = C(n+2, 3)
- S(n) = Sum[n^1] = n*(n+1)*(n+2)*(n+3)/4! = C(n+3, 4)
- 1st difference is r2 series
- 2nd difference is r1 series
- 3nd difference is common difference d=1
Q5. What is r4 series ?
A5. Answer
- r4 series : 1, 5, 15, 35, 70, ....
- T(n) = nth term = n*(n+1)*(n+2)*(n+3)/4! = C(n+3, 4)
- S(n) = Sum[n^1] = n*(n+1)*(n+2)*(n+3)*(n+4)/5! = C(n+4, 5)
- 1st difference is r3 series
- 2nd difference is r2 series
- 3rd difference is r1 series
- 4th difference is common difference d=1
Q6. Prove that Sum[n*(n+1)/2] = n*(n+1)*(n+2)/6
A6. Proof
- Sum[n*(n+1)/2] = Sum[n^2]+Sum[n/2]
- = (n*(n+1)*(2*n+1)/6 + n*(n+1)/2)/2
- = n*(n+1)*(2*n+1)/12 + n*(n+1)/4
- = n*(n+1)*(2*n+1)/3+1)/4
- = n*(n+1)*(2*n+4)/12
- = n*(n+1)*(n+2)/3! where 3!=1*2*3=6
- = C(n+2, 3) the coefficients of binomial expansion
Q6. Prove that Sum[n*(n+1)*(n+2)/3!] = n*(n+1)*(n+2)*(n+3)/4!
A6. Proof
- Sum[n*(n+1)*(n+2)/6] = Sum[n^3/6]+Sum[3*n^2/6]+Sum[2*n/6]
- = (n*(n+1)/2)/6 + n*(n+1)*(2*n+1)/12 + n*(n+1)/12
- = n*(n+1)*[(n*(n+1)/24 + (2*n+1)/12 +1/12]
- = n*(n+1)*(n^2+5*n+6)/24
- = n*(n+1)*(n+2)*(n+3)/4! where 4!=1*2*3*4
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