Use mathematical induction to prove the following for every positive integer $n$.
$\sum _ { i = 1 } ^ { n } \frac { 8 } { ( 2 i - 1 ) ( 2 i + 1 ) } = \frac { 8 n } { 2 n + 1 }$

Name: Use mathematical induction to prove the following for every positive integer $n$. $\sum _ { i = 1 } ^ { n } \frac { 8 } { ( 2 i - 1 ) ( 2 i + 1 ) } = \frac { 8 n } { 2 n + 1 }$
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Question

Quizplus · Accepted Answer

The Answer is i. Show true for

n = 1

\begin{aligned}\frac { 8 } { ( 2 \cdot 1 - 1 ) ( 2 \cdot 1 + 1 ) } & = \frac { 8 \cdot 1 } { 2 \cdot 1 + 1 } \\frac { 8 } { 3 } & = \frac { 8 } { 3 } \1 & = 1\end{aligned}

ii. Assume true for

n = k

\sum _ { i = 1 } ^ { k } \frac { 8 } { ( 2 i - 1 ) ( 2 i + 1 ) } = \frac { 8 k } { 2 k + 1 }

iii. Prove true for

n = k + 1

\begin{aligned}\sum _ { i = 1 } ^ { k + 1 } \frac { 8 } { ( 2 i - 1 ) ( 2 i + 1 ) } & = \sum _ { i = 1 } ^ { k } \frac { 8 } { ( 2 i - 1 ) ( 2 i + 1 ) } + \frac { 8 } { [ 2 ( k + 1 ) - 1 ] [ ( 2 ( k + 1 ) + 1 ) ] } \& = \frac { 8 k } { 2 k + 1 } + \frac { 8 } { ( 2 k + 1 ) ( 2 k + 3 ) } \& = \frac { 8 k ( 2 k + 3 ) + 8 } { ( 2 k + 1 ) ( 2 k + 3 ) } \& = \frac { 16 k ^ { 2 } + 24 k + 8 } { ( 2 k + 1 ) ( 2 k + 3 ) } \& = \frac { 16 k ^ { 2 } + 16 k + 8 k + 8 } { ( 2 k + 1 ) ( 2 k + 3 ) } \& = \frac { 16 k ( k + 1 ) + 8 ( k + 1 ) } { ( 2 k + 1 ) ( 2 k + 3 ) } \& = \frac { 8 ( k + 1 ) ( 2 k + 1 ) } { ( 2 k + 1 ) ( 2 k + 3 ) } \& = \frac { 8 ( k + 1 ) } { 2 ( k + 1 ) + 1 }\end{aligned}

Use Mathematical Induction to Prove the Following for Every Positive nnn

Use Mathematical Induction to Prove the Following for Every Positive $n$