Maths and Computing:

 menu
 
Home
Binary
Books
Boolean
Calculators
Coding
Computing
Encoding
Encryption
Mathematics
Series
     

Sequences and Series

Sequences and Series: Sum of 1s

Sum of 1 n times

rth term: 1

n

1

Σ

r = 1

=

1 + 1 + 1 + 1 + … + 1

n

=

n

Derivation of the Formula

n

1

Σ

r = 1

=

1 + 1 + 1 + 1 + … + 1

=

n × 1

=

n

Proof by Induction

n

1

Σ

r = 1

=

n

Check that the formula is correct when n = 1

LHS:

1

1

Σ

r = 1

=

1

RHS:

n

=

1

Assume that the formula is true for n = k

1

1

Σ

k + 1 = 1

=

1

1

Σ

k = 1

+ 1

=

k

+ 1

=

(k + 1)

If the formula is true for n = k then it is also true for n = (k + 1)


If the formula is true for n = k then it is true for n = (k + 1) and as it is true for n = 1, then by using mathematical induction the statement is true for all positive integers


Sum of the first n natural numbers

rth term: r

n

r

Σ

r = 1

=

1 + 2 + 3 + 4 + … + n

=

n(n+1)
2

Derivation of the Formula

Let S

=

n

r

Σ

r = 1

S

=

1 + 2 + 3 + … + (n-2) + (n-1) + n

S

=

n + (n-1) + (n-2) + … + 3 + 2 + 1

2S

=

[1 + n] + [2 + (n-1)] + [3 + (n-2)] + … + [(n-2) + 3] + [(n-1) + 2] + [n + 1]

=

n × (n + 1)

S

=
n(n+1)
2

Proof by Induction

The formula for the sum of the first n natural number

n

r

Σ

r = 1

=
n(n+1)
2

Check that the formula is correct when n = 1

LHS:

1

r

Σ

r = 1

=

1

RHS:
1(1+1)
2

=

1

Assume that the formula is true for n = k

1

r

Σ

k + 1 = 1

=

1

r

Σ

k = 1

+ (k+1)

=
k(k+1)
2

+ (k+1)

=
k(k+1) + 2(k+1)
2

=
(k+1)(k+2)
2

=
(k+1)((k+1)+1)
2

If the formula is true for n = k then it is also true for n = (k + 1)


If the formula is true for n = k then it is true for n = (k + 1) and as it is true for n = 1, then by using mathematical induction the statement is true for all positive integers


Sum of the first n square numbers

rth term: r2

n

r2

Σ

r = 1

=

1 + 4 + 9 + 16 + … + n2

=
n(n+1)(2n+1)
6

Derivation of the Formula

Let S =

n

r2

Σ

r = 1

n

((r + 1)3 - r3)

Σ

r = 1

=

(23 - 13) + (33 - 23) + … + (n3 - (n-1)3) + ((n+1)3 - n3)

=

(n+1)3 - 1

=

n3 + 3n2 + 3n

n

((r + 1)3 - r3)

Σ

r = 1

=

n

(3r2 + 3r + 1)

Σ

r = 1

=

3S +

3n(n+1)
2

+ n

=

3S +

3n(n+1) + 2n
2

n3 + 3n2 + 3n

=

3S +

3n2 + 5n
2

2n3 + 6n2 + 6n

=

6S + 3n2 + 5n

6S

=

2n3 + 6n2 + 6n - 3n2 - 5n

=

2n3 + 3n2 + n

=

n(n+1)(2n+1)

S

=
n(n+1)(2n+1)
6

Proof by Induction

The formula for the sum of the first n squares number

n

r2

Σ

r = 1

=
n(n+1)(2n+1)
6

Check that the formula is correct when n = 1

LHS:

n

r2

Σ

r = 1

=

1

RHS:

n

=

1

Assume that the formula is true for n = k

k+1

r2

Σ

r = 1

=

k

r2

Σ

r = 1

+ (k+1)2

=
k(k+1)(2k+1)
6

+ (k+1)2

=
k(k+1)(2k+1) + 6(k+1)2
6

=
(k+1)[k(2k+1) +6(k+1)]
6

=
(k+1)[2k2 + 7k + 6]
6

=
(k+1)(k + 2)(2k + 3)
6

=
(k+1)((k + 1) + 1)(2(k + 1) + 1)
6

If the formula is true for n = k then it is also true for n = (k + 1)


If the formula is true for n = k then it is true for n = (k + 1) and as it is true for n = 1, then by using mathematical induction, the formula is true for all positive integers


Sum of the first n cubic numbers

rth term: r3

n

r3

Σ

r = 1

=

1 + 8 + 27 + 64 + … + n3

=
n2(n+1)2
4

Derivation of the Formula

Let S =

n

r3

Σ

r = 1

n

((r + 1)4 - r4)

Σ

r = 1

= (24 - 14) + (34 - 24) + … + (n4 - (n-1)4) + ((n+1)4 - n4)

= (n+1)4 - 1

= n4 + 4n3 + 6n2 + 4n

n

((r + 1)4 - r4)

Σ

r = 1

=

n

(4r3 + 6r2 + 4r + 1)

Σ

r = 1

=

4S +

6n(n+1)(2n+1)
6

+
4n(n+1)
2

+ n

=

4S + n(n+1)(2n+1) + 2n(n+1) + n

=

4S + 2n3 + 5n2 + 4n

n4 + 4n3 + 6n2 + 4n

=

4S + 2n3 + 5n2 + 4n

4S

=

n4 + 2n3 + n2

=

n2(n + 1)2

S

=
n2(n+1)2
4

Proof by Induction

The formula for the sum of the first n cubic number

n

r2

Σ

r = 1

=
n2(n+1)2
4

Check that the formula is correct when n = 1

LHS:

n

r3

Σ

r = 1

=

1

RHS:

n

=

1

Assume that the formula is true for n = k

k+1

r2

Σ

r = 1

=

k

r3

Σ

r = 1

+ (k+1)2

=
k2(k+1)2
4

+ (k+1)3

=
k2(k+1)2 + 4(k+1)3
4

=
(k+1)2[k2 +4(k+1)]
4

=
(k+1)2[k2 +4k+4)]
4

=
(k+1)2(k + 2)2
4

=
(k+1)2((k + 1) + 1)2
4

If the formula is true for n = k then it is also true for n = (k+1)


If the formula is true for n = k then it is true for n = (k + 1) and as it is true for n = 1, then by using mathematical induction, the formula is true for all positive integers


Sums of Polynomials

Expression for

n

ar3 + br2 + cr + d

Σ

r = 1

a:

b:

c:

d:

n

(r3 + r2 + r + 1)

Σ

r = 1

=
3n4 + 10n3 + 15n2 + 20n
12

=
n(3n3 + 10n2 + 15n + 20)
12
Sum of 1 n times

n

1

Σ

r = 1

=

n
Sum of the first n numbers

n

r

Σ

r = 1

=
n(n+1)
2
Sum of the first n square numbers

n

r2

Σ

r = 1

=
n(n+1)(2n+1)
6
Sum of the first n cube numbers

n

r3

Σ

r = 1

=
n2(n+1)2
4

n

(r3 + r2 + r + 1)

Σ

r = 1

=

n

r3

Σ

r = 1

 +

n

r2

Σ

r = 1

 +

n

r

Σ

r = 1

 +

n

1

Σ

r = 1

=
n2(n+1)2
4
 +
n(n+1)(2n+1)
6
 +
n(n+1)
2
 +

n

=
n(3n3 + 10n2 + 15n + 20)
12

=
n(3n3 + 10n2 + 15n + 20)
12

Contents Copyright 2026 Andy Abraham

Please email comments to info@mathsandcomputing.co.uk