Answer to Question #135852 in Algebra for Yolanda

Question

using mathematical induction prove that (n+1)^2<2n^2 for all natural numbers n=3 or greater than 3

Answer 1

Step 1. Prove the result is true for $n=3$:

Left Side: $(n+1)^2$ Right Side: $2⋅n^2$

Left Side: $(3+1)^2$ Right Side: $2⋅3^2$

Left Side: $4^2$ Right Side: $2⋅9$

Left Side: $16$ Right Side: $18$

Left Side $<$ Right Side

Hence the result is true for $n=3$.

Step 2. Assume the result is true for $n=k$, where $k$ is a positive integer $\geqslant3$:

$(k+1)^2<2k^2$

Step 3. Prove the result is true for $n=k+1$:

$(k+1+1)^2<2⋅(k+1)^2$

$(k+2)^2<2⋅(k+1)^2$

$k^2+4k+4<2⋅(k^2+2k+1)$

$k^2+4k+4<2k^2+4k+2$

$2<k^2$ by $k\geqslant3$

So, $(k+1+1)^2<2⋅(k+1)^2$ holds by $n=k+1$, therefore

$(n+1)^2<2n^2$ is true for $n\geqslant3$.