Prove by mathematical induction that

2 + 4 + 6 + 8 + ... + 2n = n^2 + n, for all n = 1.

Question

Prove by mathematical induction that

2 + 4 + 6 + 8 + ... + 2n = n^2 + n, for all n >= 1.

anonymous · Answer

Let the aforementioned be the induction hypothesis.  Now consider $n=1$.  Thus, by the induction hypothesis yields the following.$$2=(1)^2+1$$This is obviously true.  The induction step is then as follows.  Now, let us add $2(n+1)$ (an operation which is valid for all $n\ge1$) to both sides of the induction hypothesis which we will follow with some slight algebraic rearrangement.$$\begin{align} 2+4+6+8+\cdots+2n+2(n+1)&=n^2+n+2(n+1) \ 2+4+6+8+\cdots+2(n+1)&=(n+1)^2+(n+1) \end{align}$$This demonstrates the induction hypothesis holds for all $n+1$ when $n\ge 1$ and thus since the induction hypothesis holds for $n=1$, the proposed equality holds for all $n\ge1$.

anonymous · Answer

Thanks!