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- Homework Statement
- Prove by induction on ##n## that when ##x>0##,

- Relevant Equations
- ##(1+x)^n\ge 1+nx+\frac{n(n-1)}{2}x^2## for all positive intergers ##n##.

The statements holds for the case ##n=1##

\begin{align} 1+x \leq & 1+x+\frac{1}{2}\cdot 0\cdot x^2\\

=&1+x \end{align}

Assume the statement holds true for ##n=k##

$$(1+x)^k\geq 1+kx+\frac{k(k-1)}{2} x^2$$

Then, for ##n=k+1##, we have the following

\begin{align} (1+x)^k\cdot (1+x)\geq& 1+kx+\frac{k(k-1)}{2} x^2+x+kx^2\\

(1+x)^{k+1}\geq& 1+(k+1)x+\frac{(k+1)k}{2} x^2 \end{align}

Which means the statement holds for all positive integers ##n##.

I am unsure about the term ##+kx^2## in the second to last line. I computed it by working backwards from the conclusion and solving for ##A## $$k(k-1)+A=(k+1)k$$But I do not know how to obtain the term working in the forward direction.

\begin{align} 1+x \leq & 1+x+\frac{1}{2}\cdot 0\cdot x^2\\

=&1+x \end{align}

Assume the statement holds true for ##n=k##

$$(1+x)^k\geq 1+kx+\frac{k(k-1)}{2} x^2$$

Then, for ##n=k+1##, we have the following

\begin{align} (1+x)^k\cdot (1+x)\geq& 1+kx+\frac{k(k-1)}{2} x^2+x+kx^2\\

(1+x)^{k+1}\geq& 1+(k+1)x+\frac{(k+1)k}{2} x^2 \end{align}

Which means the statement holds for all positive integers ##n##.

I am unsure about the term ##+kx^2## in the second to last line. I computed it by working backwards from the conclusion and solving for ##A## $$k(k-1)+A=(k+1)k$$But I do not know how to obtain the term working in the forward direction.