Question 1
a1 = 1∙2 = 2
a2 = 2∙3 = 6
a3 = 3∙4 = 12
a4 = 4∙5 = 20
a5 = 5∙6 = 30
a6 = 6∙7 = 42
an = n(n+1) = n2+n
Question 2
A)
S1 = a1 = 2
S2 = a1+ a2 =8
S3 = a1+…+ a3 = 20
S4 = a1+…+ a4 = 40
S5 = a1+…+ a5 = 70
S6 = a1+…+ a6 = 112
Sk = a1+…+ ak
B) Sn =
=
=
=
=
∴ Sn =
C) 1∙2+2∙3+3∙4+…+n(n+1) =
i) n=1
1∙2 =
2 = 2
ii) n=k
1∙2+2∙3+3∙4+…+k(k+1) =
Add (k+1)(k+2) both side
1∙2+2∙3+3∙4+…+k(k+1)+(k+1)(k+2) = + (k+1)(k+2)
1∙2+2∙3+3∙4+…+k(k+1)+(k+1)(k+2) =
1∙2+2∙3+3∙4+…+k(k+1)+(k+1)(k+2) =
iii) n=k+1
1∙2+2∙3+3∙4+…+(k+1)(k+2) =
D) 12+22+32+42+…+n2
=
=
= -
=
=
∴ =
Question 3
A)
T1 =1∙2∙3 = 6
T2 =1∙2∙3+2∙3∙4 = 6+24 = 30
T3 =1∙2∙3+…+3∙4∙5 = 30+60 = 90
T4 =1∙2∙3+…+4∙5∙6 = 90+120 = 210
T5 =1∙2∙3+…+5∙6∙7 = 210+210 = 420
T6 =1∙2∙3+…+6∙7∙8 = 420+336 = 756
Tk =1∙2∙3+…+n(n+1)(n+2)
B) Tn =
=
=
=
=
∴Tn =
C) 1∙2∙3+2∙3∙4+3∙4∙5+…+n(n+1)(n+2) =
i) n=1
1∙2∙3 =
6 = 6
ii) n=k
1∙2∙3+2∙3∙4+3∙4∙5+…+k(k+1)(k+2) =
Add (k+1)(k+2)(k+3) both side
1∙2∙3+2∙3∙4+3∙4∙5+…+k(k+1)(k+2)+(k+1)(k+2)(k+3) = + (k+1)(k+2)(k+3)
1∙2∙3+2∙3∙4+3∙4∙5+…+k(k+1)(k+2)+(k+1)(k+2)(k+3) =
1∙2∙3+2∙3∙4+3∙4∙5+…+k(k+1)(k+2)+(k+1)(k+2)(k+3) =
iii) n=k+1
1∙2∙3+2∙3∙4+3∙4∙5+…+(k+1)(k+2)(k+3) =
D) 13+23+33+43+…+n3
=
=
=
=
=
=
∴ =
Question 4
A)
U1 =1∙2∙3∙4 = 24
U2 =1∙2∙3∙4+2∙3∙4∙5 = 24+120 = 144
U3 =1∙2∙3∙4+…+3∙4∙5∙6 = 144+360 = 504
U4 =1∙2∙3∙4+…+4∙5∙6∙7 = 504+840 = 1344
U5 =1∙2∙3∙4+…+5∙6∙7∙8 = 1344+1680 = 3024
U6 =1∙2∙3∙4+…+6∙7∙8∙9 = 3024+3024 = 6048
Uk =1∙2∙3∙4+…+n(n+1)(n+2)(n+3)
B)
Un =
∴Un
C)
1∙2∙3∙4+2∙3∙4∙5+3∙4∙5∙6+…+n(n+1)(n+2)(n+3) =
i) n=1
1∙2∙3∙4 =
24 = 24
ii) n=k
1∙2∙3∙4+2∙3∙4∙5+3∙4∙5∙6+…+k(k+1)(k+2)(k+3) =
Add (k+1)(k+2)(k+3)(k+4) both side
1∙2∙3∙4+2∙3∙4∙5+3∙4∙5∙6+…+k(k+1)(k+2)(k+3)+ (k+1)(k+2)(k+3)(k+4) = + (k+1)(k+2)(k+3)(k+4)
1∙2∙3∙4+2∙3∙4∙5+3∙4∙5∙6+…+k(k+1)(k+2)(k+3)+ (k+1)(k+2)(k+3)(k+4) =
1∙2∙3∙4+2∙3∙4∙5+3∙4∙5∙6+…+k(k+1)(k+2)(k+3)+ (k+1)(k+2)(k+3)(k+4) =
iii) n=k+1
1∙2∙3∙4+2∙3∙4∙5+3∙4∙5∙6+…+(k+1)(k+2)(k+3)(k+4) =
D) 14+24+34+44+…+n4
=
=
∴
Question 5
Use of the Pascal’s triangle
1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
…… And so on
By using Pascal’s triangle, we can get the results with binomial theorem.
We could use this triangle to get and put those coefficients to
(n+1)k+1 –nk+1 = ank+bnk-1+cnk-2+…1
n=1
2k+1 –1k+1 = a∙1k+b∙1k-1+c∙1k-2+…1
n=2
3k+1 –2k+1 = a∙2k+b∙2k-1+c∙2k-2+…1
n=m
(m +1)k+1 – m k+1 = ank+bnk-1+cnk-2+…1
= a(1 k +2 k +3 k +…+ m k ) + b(1 k-1 +2 k-1 +3 k-1 +…+ m k-1)+ …+ m
∴ = 1 k+2 k+3 k+4 k +…+n k ={(m+1) k+1–1–b(1 k-1 +2 k-1 +3 k-1 +…+ m k-1)+…–m}
Those a, b, c, d represents the coefficients. We can simply get the coefficient from the Pascal’s triangle.
