Determining Initial Values | Principles of Mathematical Induction

Prove \( 1+3+5+\cdots+(2n+1) = (n+1)^2 \).

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Step 1

Show it is true for \( n=0 \) by mathematical induction.

\( \begin{align} &\text{LHS} = 2 \times 0 +1 = 1 \\ &\text{RHS} = (0+1)^2 = 1 \\ &\text{LHS} = \text{RHS} \\ &\text{Therefore, it is true for } n=0 \end{align} \)

Step 2

Assume that it is true for \( n=k \).

\( \text{That is, } 1+3+5+ \cdots +(2k+1) = (k+1)^2 \)

Step 3

Show it is true for \( n=k+1 \).

\( \begin{align} \text{That is, } &1+3+5+\cdots+(2k+1)+(2k+3)=(k+2)^2 \\ \text{LHS} &= 1+3+5+\cdots+(2k+1)+(2k+3) \\ &=(k+1)^2+(2k+3) \\ &= k^2+2k+1+2k+3 \\ &= k^2+4k+4 \\ &=(k+2)^2 \\ &=\text{RHS} \\ \text{Therefore } &\text{it is true for } n=k+1. \\ \text{Therefore } &\text{the statement is true for all integers } n\ge 0. \end{align} \)

 

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