Use Mathematical Induction to Prove That for All Integers$n \geq 1$

Question

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The Answer is

\underline { ext { Proof (by mathematical induction) } }

: Let the property

P ( n )

be the equation

4 + 8 + 12 + \cdots + 4 n = 2 n ^ { 2 } + 2 n . \quad \leftarrow P ( n )

Note that

4 + 8 + 12 + \cdots + 4 n = \sum _ { i = 1 } ^ { n } 4 i

Show that

P ( 1 )

is true:

P ( 1 )

is true because the left-hand side is

\sum _ { i = 1 } ^ { 1 } 4 i = 4 \cdot 1 = 4

and the right-hand side is

2 \cdot 1 ^ { 2 } + 2 \cdot 1 = 4

also.

Show that for all integers

k \geq 1

, if

P ( k )

is true then

P ( k + 1 )

is true: Let

k

be any integer with

k \geq 1

, and suppose that

4 + 8 + 12 + \cdots + 4 k = 2 k ^ { 2 } + 2 k . \quad \leftarrow \quad \begin{array} { l } P ( k ) \ ext { inductive hypothesis }\end{array}

We must show that

4 + 8 + 12 + \cdots + 4 ( k + 1 ) = 2 ( k + 1 ) ^ { 2 } + 2 ( k + 1 ) . \quad \leftarrow P ( k + 1 )

Now the left-hand side of

P ( k + 1 )

is

\begin{aligned} 4 + 8 + 12 + \cdots + 4 ( k + 1 ) = & 4 + 8 + 12 + \cdots + 4 k + 4 ( k + 1 ) \ = & ext { by making the next-to-last term explicit } \ & \left( 2 k ^ { 2 } + 2 k ight) + 4 ( k + 1 ) \ & ext { by inductive hypothesis } \ & = \begin{array} { c } 2 k ^ { 2 } + 6 k + 4 \ ext { by multiplying out and combining like terms. } \end{array} \end{aligned}

And the right-hand side of

P ( k + 1 )

is

2 ( k + 1 ) ^ { 2 } + 2 ( k + 1 ) = 2 \left( k ^ { 2 } + 2 k + 1 ight) + 2 k + 2 = 2 k ^ { 2 } + 4 k + 2 + 2 k + 2 = 2 k ^ { 2 } + 6 k + 4 .

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 $n \geq 1$