‌‌
Summation Formulas
‌‌
Every calculus course uses summation formulas.  In most classes the fomulas are either not proven or are proven by mathematical induction.  But this gives no indication of how the formulas can be derived.  One method of deriving the formula for ‌
n
N
i=1
i
‌2
‌ is to assume that the sum can be written in the form a
n
‌3
‌ + b
n
‌2
‌ + cn + d and plugging in values on n.  Using n equals 1, 2, 3 and 4 yields the system of equations:  ‌
A
B
B
B
C
H
a
+
b
+
c
+
d
=
1
8a
+
4b
+
2c
+
d
=
3
27a
+
9b
+
3c
+
d
=
6
64a
+
16b
+
4c
+
d
=
10
.‌  Solving the system yields the coefficients.  However, this method just begs the question of how do you know there is a solution of the given form.  The following is an interesting way to derive the formula for ‌
n
N
i=1
i
‌2
‌:

n+1
N
i=1
i
‌3
‌ = ‌
n
N
i=0
‌ (i + 1
)
‌3
‌ = ‌
n
N
i=0
‌ (
i
‌3
‌ + 3
i
‌2
‌ + 3i + 1)
           = ‌
n
N
i=0
i‌
‌3
‌ + 3‌
n
N
i=0
i‌
‌2
‌ + 3‌
n
N
i=0
‌ i + ‌
n
N
i=0
‌ 1
           = ‌
n
N
i=1
i‌
‌3
‌ + 3‌
n
N
i=1
i
‌2
‌ + 3‌
n
N
i=1
‌ i + n + 1
 ‌
n+1
N
i=1
i
‌3
‌ J‌ ‌
n
N
i=1
i
‌3
‌ = 3‌
n
N
i=1
i
‌2
‌ + 3‌
n
N
i=1
‌ i + n + 1
(n+1
)
‌3
‌ = 3‌
n
N
i=1
i
‌2
‌ + 3‌
n
N
i=1
‌ i + n + 1
(n + 1
)
‌3
‌ J‌ (n + 1) J‌ 3‌
n
N
i=1
‌ i = 3‌
n
N
i=1
i
‌2
3‌
n
N
i=1
i‌
‌2
‌ = (n + 1
)‌
‌3
‌ J‌ (n + 1) J‌ 3‌
n(n + 1)
2
            = (n + 1)[(n + 1
)
‌2
‌ J‌ 1] J‌ ‌
3n(n + 1)
2
            = (n + 1)(n)(n + 2) J‌ 
3n(n + 1)
2
            = n(n + 1)[n + 2 J‌ ‌
3
2
‌]
            = n(n + 1)‌
2n + 1
2
n
N
i=1
i
‌2
‌ = ‌
n(n + 1)(2n + 1)
6

Now let's consider the general case:

n+1
N
i=1
i
‌m+1
‌ = ‌
n
N
i=0
‌ ‌(‌
i + 1)
‌m+1
‌ = ‌
n
N
i=0
‌ ‌
m+1
N
j=0
‌ ‌
A
B
C
m+1
j
K
L
M
 ‌
i
‌j
           = ‌
m+1
N
j=0
‌ ‌
A
B
C
m+1
j
K
L
M
 ‌‌
n
N
i=0
i
‌j
           = n + 1 + (m + 1) ‌
n
N
i=0
i
‌m
‌ + ‌
n
N
i=0
i
‌m+1
‌ + ‌
m-1
N
j=1
‌ ‌
A
B
C
m+1
j
K
L
M
 ‌‌
n
N
i=0
i
‌j
           = n + 1 + (m + 1)‌
n
N
i=1
i
‌m
‌ + ‌
n
N
i=1
i
‌m+1
‌ + ‌
m-1
N
j=1
‌ ‌
A
B
C
m+1
j
K
L
M
n
N
i‌=1
i
‌j

n+1
N
i=1
i
‌m+1
‌ J‌ ‌
n
N
i=1
i
‌m+1
‌ = n + 1 + (m + 1) ‌
n
N
i=1
i
‌m
‌ + ‌
m-1
N
j=1
‌ ‌
A
B
C
m+1
j
K
L
M
 ‌‌
n
N
i=1
i
‌j
(n + 1
)
‌m+1
‌ - (n + 1) = (m + 1) ‌
n
N
i=1
i
‌m
‌ + ‌
m-1
N
j=1
‌ ‌
A
B
C
m+1
j
K
L
M
 ‌‌
n
N
i=1
i
‌j
(m + 1) ‌
n
N
i=1
i
‌m
‌ = (n + 1)[(n + 1
)
‌m
‌ J‌ 1] J‌ ‌
m-1
N
j=1
‌ ‌
A
B
C
m+1
j
K
L
M
‌ 
n
N
i=1
‌ ‌
i
‌j
and thus,
n
N
i=1
i
‌m
‌ = ‌(n + 1)[(n + 1‌
)
‌m
‌ J‌ 1] J‌ ‌
m-1
N
j=1
‌ ‌
A
B
C
m+1
j
K
L
M
 ‌‌
n
N
i=1
‌ ‌
i
‌j
‌.


Some formulas derived from this recursive formula are given below:
n
N
i=1
i
‌2
‌ = ‌
2n3 + 3n2 + n
6
= ‌
n(n + 1)(2n + 1)
6
n
N
i=1
i
‌3
‌ = ‌
n4 + 2n3 + n2
4
 ‌
= ‌‌
n
‌2
‌(n + 1‌
)
‌2
4
n
N
i=1
i
‌4
‌ = ‌
6n5 + 15n4 + 10n3 - n
30
= ‌
n(n + 1)(2n + 1)(3‌
n
‌2
‌ + 3n J‌ 1)
30
n
N
i=1
i
‌5
‌ = ‌‌‌
2n6 + 6n5 + 5n4 - n2
12
= ‌‌
n
‌2
‌(n + 1‌
)
‌2
‌(2‌
n
‌2
‌ + 2n J‌ 1)
12
n
N
i=1
i
‌6
‌ = ‌
6n7 + 21n6 + 21n5 - 7n3 + n
42
=
n(n + 1)(2n + 1)(3‌
n
‌4
‌ + 6‌
n
‌3
‌ J‌ 3n + 1)
42
n
N
i=1
i
‌7
‌ = ‌‌‌
3n8 + 12n7 + 14n6 - 7n4 + 2n2
24
‌ =
‌ 
n
‌2
‌(n + 1‌
)
‌2
‌(3‌
n
‌4
‌ + 6‌
n
‌3
‌ J‌ ‌
n
‌2
‌ J‌ 4n + 2)
24
n
N
i=1
i
‌8
‌ = ‌‌‌‌
10n9 + 45n8 + 60n7 - 42n5 + 20n3 - 3n
90
= ‌
n(n +1)(2n + 1)(5‌
n
‌6
‌ + 15‌
n
‌5
‌ + 5‌
n
‌4
‌ J‌ 15‌
n
‌3
‌ J‌ ‌
n
‌2
‌ + 9n J‌ 3)
90
n
N
i=1
i
‌9
‌ = ‌‌‌
2n
10
 + 10n9 + 15n8 - 14n6 + 10n4 - 3n2
20
= ‌‌
n
‌2
‌(n + 1‌
)
‌2
‌(‌
n
‌2
‌ + n J‌ 1)(2‌
n
‌4
‌+ 4‌
n
‌3
‌ J‌ ‌
n
‌2
‌ J‌ 3n + 3)
20
 
n
N
i=1
i
‌10
‌ = ‌
6n
11 
+ 33n
10 
+ 55n9 - 66n7 + 66n5 - 33n3 + 5n
66
= ‌
n(n + 1)(2n + 1)(‌
n
‌2
‌ + n J‌ 1)(3‌
n
‌6
‌ + 9‌
n
‌5
‌ + 2‌
n
‌4
‌ J‌ 11‌
n
‌3
‌ + 3‌
n
‌2
‌ + 10n J‌ 5)
66