Rule

Formulas for Special Series

When a series has many terms, it can be tedious to calculate its sum. However, there are formulas for special series that simplify calculations. The table shows the formulas for some special series.

Special Series	Formula
Sum of $n$ terms of $1 .$	$i = 1 \sum n 1 = n$
Sum of first $n$ positive integers.	$i = 1 \sum n i = \frac{n ( n + 1 )}{2}$
Sum of squares of first $n$ positive integers.	$i = 1 \sum n i^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}$
Sum of cubes of first $n$ positive integers.	$i = 1 \sum n i^{3} = [\frac{n ( n + 1 )}{2}]^{2}$

Proof

Sum of

n

Terms of

1

Adding the same number

n

times is the same as multiplying the number by

n .

In this case, since the number is

1,

the result of this operation is

1 \cdot n,

which equals

n .

i = 1 \sum n 1 = n times 1 + 1 + 1 + \dots + 1 = n \cdot 1 = n

Proof

Sum of First

n

Positive Integers

Mathematical induction will be used to prove the following statement.

i = 1 \sum n i = \frac{n ( n + 1 )}{2}

The first step in an inductive proof is to show that the equation holds true for

n = 1 .

i = 1 \sum n i = \frac{n ( n + 1 )}{2}

Substitute

$n = 1$

i = 1 \sum 1 i = ? \frac{1 ( 1 + 1 )}{2}

Rewrite

Rewrite $i = 1 \sum 1 i$ as $1$

1 = ? \frac{1 ( 1 + 1 )}{2}

AddTerms

Add terms

1 = ? \frac{1 ( 2 )}{2}

IdPropMult

Identity Property of Multiplication

1 = ? \frac{2}{2}

QuotOne

$\frac{a}{a} = 1$

1 = 1 ✓

The statement is true for

n = 1 .

Now, assume that the statement is true for some positive integer

k .

i = 1 \sum k i = \frac{k ( k + 1 )}{2} ✓

In the final step, the aim is to show that the statement is true for

n = k + 1 .

To do so, the above equation will be manipulated using the Properties of Equality.

i = 1 \sum k i = \frac{k ( k + 1 )}{2}

AddEqn

$LHS + (k + 1) = RHS + (k + 1)$

i = 1 \sum k i + (k + 1) = \frac{k ( k + 1 )}{2} + (k + 1)

Rewrite

Rewrite $i = 1 \sum k i + (k + 1)$ as $i = 1 \sum k + 1 i$

i = 1 \sum k + 1 i = \frac{k ( k + 1 )}{2} + (k + 1)

Simplify right-hand side

NumberToFrac

$a = \frac{2 \cdot a}{2}$

i = 1 \sum k + 1 i = \frac{k ( k + 1 )}{2} + \frac{2 ( k + 1 )}{2}

AddFrac

Add fractions

i = 1 \sum k + 1 i = \frac{k ( k + 1 ) + 2 ( k + 1 )}{2}

FactorOut

Factor out $(k + 1)$

i = 1 \sum k + 1 i = \frac{( k + 1 ) ( k + 2 )}{2}

WriteSum

Write as a sum

i = 1 \sum k + 1 i = \frac{( k + 1 ) ( k + 1 + 1 )}{2}

The left-hand side of the above equation is the sum of the first

k + 1

positive integers. The right-hand side is the expression obtained when substituting

k + 1

for

n .

Therefore, the statement holds true for all positive integers.

Proof

Sum of Squares of First

n

Positive Integers

The following statement will be proved using mathematical induction.

i = 1 \sum n i^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}

For the first step, it should be shown that the equation holds true for

n = 1 .

i = 1 \sum n i^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}

Substitute

$n = 1$

i = 1 \sum 1 i^{2} = ? \frac{1 ( 1 + 1 ) ( 2 ( 1 ) + 1 )}{6}

Evaluate left-hand side

Rewrite

Rewrite $i = 1 \sum 1 i^{2}$ as $1^{2}$

1^{2} = ? \frac{1 ( 1 + 1 ) ( 2 ( 1 ) + 1 )}{6}

BaseOne

$1^{a} = 1$

1 = ? \frac{1 ( 1 + 1 ) ( 2 ( 1 ) + 1 )}{6}

Evaluate right-hand side

IdPropMult

Identity Property of Multiplication

1 = ? \frac{1 ( 1 + 1 ) ( 2 + 1 )}{6}

AddTerms

Add terms

1 = ? \frac{1 ( 2 ) ( 3 )}{6}

Multiply

1 = ? \frac{6}{6}

QuotOne

$\frac{a}{a} = 1$

1 = 1 ✓

The statement is true for

n = 1 .

Now, assume that the statement is true for some positive integer

k .

i = 1 \sum k i^{2} = \frac{k ( k + 1 ) ( 2 k + 1 )}{6} ✓

Next, show that the statement is true for

n = k + 1 .

To do so, start by adding

(k + 1)^{2}

to both sides of the above equation.

i = 1 \sum k i^{2} = \frac{k ( k + 1 ) ( 2 k + 1 )}{6}

AddEqn

$LHS + (k + 1)^{2} = RHS + (k + 1)^{2}$

i = 1 \sum k i^{2} + (k + 1)^{2} = \frac{k ( k + 1 ) ( 2 k + 1 )}{6} + (k + 1)^{2}

Rewrite

Rewrite $i = 1 \sum k i^{2} + (k + 1)^{2}$ as $i = 1 \sum k + 1 i^{2}$

i = 1 \sum k + 1 i^{2} = \frac{k ( k + 1 ) ( 2 k + 1 )}{6} + (k + 1)^{2}

Add the terms on the right-hand side, then factor the numerator of the fraction.

i = 1 \sum k + 1 i^{2} = \frac{k ( k + 1 ) ( 2 k + 1 )}{6} + (k + 1)^{2}

Simplify right-hand side

NumberToFrac

$a = \frac{6 \cdot a}{6}$

i = 1 \sum k + 1 i^{2} = \frac{k ( k + 1 ) ( 2 k + 1 )}{6} + \frac{6 ( k + 1 ) ^{2}}{6}

AddFrac

Add fractions

i = 1 \sum k + 1 i^{2} = \frac{k ( k + 1 ) ( 2 k + 1 ) + 6 ( k + 1 ) ^{2}}{6}

FactorOut

Factor out $(k + 1)$

i = 1 \sum k + 1 i^{2} = \frac{( k + 1 ) [ k ( 2 k + 1 ) + 6 ( k + 1 ) ]}{6}

Distr

Distribute $k & 6$

i = 1 \sum k + 1 i^{2} = \frac{( k + 1 ) ( 2 k ^{2} + k + 6 k + 6 )}{6}

WriteSum

Write as a sum

i = 1 \sum k + 1 i^{2} = \frac{( k + 1 ) ( 2 k ^{2} + k + 3 k + 3 k + 6 )}{6}

AddTerms

Add terms

i = 1 \sum k + 1 i^{2} = \frac{( k + 1 ) ( 2 k ^{2} + 4 k + 3 k + 6 )}{6}

FactorOut

Factor out $2 k$

i = 1 \sum k + 1 i^{2} = \frac{( k + 1 ) [ 2 k ( k + 2 ) + 3 k + 6 ]}{6}

FactorOut

Factor out $3$

i = 1 \sum k + 1 i^{2} = \frac{( k + 1 ) [ 2 k ( k + 2 ) + 3 ( k + 2 ) ]}{6}

FactorOut

Factor out $(k + 2)$

i = 1 \sum k + 1 i^{2} = \frac{( k + 1 ) ( k + 2 ) ( 2 k + 3 )}{6}

WriteSum

Write as a sum

i = 1 \sum k + 1 i^{2} = \frac{( k + 1 ) ( k + 1 + 1 ) ( 2 k + 2 + 1 )}{6}

FactorOut

Factor out $2$

i = 1 \sum k + 1 i^{2} = \frac{( k + 1 ) ( k + 1 + 1 ) [ 2 ( k + 1 ) + 1 ]}{6}

The left-hand side is the sum of the squares of the first

n

positive integers. The right-hand side is the expression obtained when substituting

k + 1

for

n .

Therefore, the statement holds for all positive integers

n .

Proof

Sum of Cubes of First

n

Positive Integers

The following statement can be proved by mathematical induction.

i = 1 \sum n i^{3} = [\frac{n ( n + 1 )}{2}]^{2}

In the first step, it should be shown that the equation holds true for

n = 1 .

i = 1 \sum n i^{3} = [\frac{n ( n + 1 )}{2}]^{2}

Substitute

$n = 1$

i = 1 \sum 1 i^{3} = ? [\frac{1 ( 1 + 1 )}{2}]^{2}

Evaluate left-hand side

Rewrite

Rewrite $i = 1 \sum 1 i^{3}$ as $1^{3}$

1^{3} = ? [\frac{1 ( 1 + 1 )}{2}]^{2}

BaseOne

$1^{a} = 1$

1 = ? [\frac{1 ( 1 + 1 )}{2}]^{2}

Evaluate right-hand side

AddTerms

Add terms

1 = ? [\frac{1 ( 2 )}{2}]^{2}

IdPropMult

Identity Property of Multiplication

1 = ? (\frac{2}{2})^{2}

QuotOne

$\frac{a}{a} = 1$

1 = ? 1^{2}

BaseOne

$1^{a} = 1$

1 = 1 ✓

The statement is true for

n = 1 .

Now, assume that the statement is true for some positive integer

k .

i = 1 \sum k i^{3} = [\frac{k ( k + 1 )}{2}]^{2} ✓

Finally, to show that the statement is true for

k + 1,

the above equation will be manipulated.

i = 1 \sum k i^{3} = \frac{k ^{2} ( k + 1 ) ^{2}}{4}

AddEqn

$LHS + (k + 1)^{3} = RHS + (k + 1)^{3}$

i = 1 \sum k i^{3} + (k + 1)^{3} = \frac{k ^{2} ( k + 1 ) ^{2}}{4} + (k + 1)^{3}

Rewrite

Rewrite $i = 1 \sum k i^{3} + (k + 1)^{3}$ as $i = 1 \sum k + 1 i^{3}$

i = 1 \sum k + 1 i^{3} = \frac{k ^{2} ( k + 1 ) ^{2}}{4} + (k + 1)^{3}

Simplify right-hand side

NumberToFrac

$a = \frac{4 \cdot a}{4}$

i = 1 \sum k + 1 i^{3} = \frac{k ^{2} ( k + 1 ) ^{2}}{4} + \frac{4 ( k + 1 ) ^{3}}{4}

AddFrac

Add fractions

i = 1 \sum k + 1 i^{3} = \frac{k ^{2} ( k + 1 ) ^{2} + 4 ( k + 1 ) ^{3}}{4}

FactorOut

Factor out $(k + 1)^{2}$

i = 1 \sum k + 1 i^{3} = \frac{( k + 1 ) ^{2} ( k ^{2} + 4 ( k + 1 ) )}{4}

Distr

Distribute $4$

i = 1 \sum k + 1 i^{3} = \frac{( k + 1 ) ^{2} ( k ^{2} + 4 k + 4 )}{4}

FacPosPerfectSquare

$a^{2} + 2 a b + b^{2} = (a + b)^{2}$

i = 1 \sum k + 1 i^{3} = \frac{( k + 1 ) ^{2} ( k + 2 ) ^{2}}{4}

ProdPowII

$a^{m} b^{m} = (a b)^{m}$

i = 1 \sum k + 1 i^{3} = \frac{[ ( k + 1 ) ( k + 2 ) ] ^{2}}{4}

WritePow

Write as a power

i = 1 \sum k + 1 i^{3} = \frac{[ ( k + 1 ) ( k + 2 ) ] ^{2}}{2 ^{2}}

$\frac{a ^{m}}{b ^{m}} = (\frac{a}{b})^{m}$

i = 1 \sum k + 1 i^{3} = [\frac{( k + 1 ) ( k + 2 )}{2}]^{2}

WriteSum

Write as a sum

i = 1 \sum k + 1 i^{3} = [\frac{( k + 1 ) ( k + 1 + 1 )}{2}]^{2}

This final equation shows that the statement is true for

n = k + 1 .

Therefore, the statement holds for all positive integers

n .

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Proof

Proof

Proof

Proof