Use mathematical induction to prove that for all integers $n \geq 3$,
$$3 + 4 + 5 + \cdots + n = \frac { ( n - 2 ) ( n + 3 ) } { 2 } .$$

Question

Quizplus · Accepted Answer

The Answer is Proof (by mathematical induction): Let the property

P ( n )

be the equation

3 + 4 + 5 + \cdots + n = \frac { ( n - 2 ) ( n + 3 ) } { 2 } . \quad \leftarrow P ( n )

Note that

3 + 4 + 5 + \cdots + n = \sum _ { i = 3 } ^ { n } i

Show that

P ( 3 )

is true:

P ( 3 )

is true because the left-hand side is

\sum _ { i = 3 } ^ { 3 } i = 3

and the righthand side is

\frac { ( 3 - 2 ) ( 3 + 3 ) } { 2 } = 3

also.
Show that for all integers

k \geq 3

, if

P ( k )

is true then

P ( k + 1 )

is true: Let

k

be any integer with

k \geq 1

, and suppose that

3 + 4 + 5 + \cdots + k = \frac { ( k - 2 ) ( k + 3 ) } { 2 } . \leftarrow \begin{array} { l } P ( k ) \ ext { inductive hypothesis }\end{array}

We must show that

3 + 4 + 5 + \cdots + ( k + 1 ) = \frac { [ ( k + 1 ) - 2 ] [ ( k + 1 ) + 3 ] } { 2 } . \leftarrow P ( k + 1 )

Now the left-hand side of

P ( k + 1 )

is

\begin{aligned}3 + 4 + 5 + \cdots + ( k + 1 ) & = 3 + 4 + 5 + \cdots + k + ( k + 1 ) \& = \frac { ( k - 2 ) ( k + 3 ) } { 2 } + ( k + 1 ) \& = \frac { ( k - 2 ) ( k + 3 ) } { ext { by making the next-to-last term explicit } } + \frac { 2 \cdot ( k + 1 ) } { 2 } \& = \frac { k ^ { 2 } + k - 6 } { 2 } + \frac { 2 k + 2 } { 2 } \& = \frac { k ^ { 2 } + 3 k - 4 } { 2 } \& = \frac { ext { by multiplying out } } { 2 } \ ext { by adding fractions and combining like termoting a common denominator }\end{aligned}

And the right-hand side of

P ( k + 1 )

is

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

Thus the left-hand and right-hand sides of

P ( k + 1 )

are equal [as was to be shown].

Use Mathematical Induction to Prove That for All Integers n≥3n \geq 3n≥3