# Mathematical Induction 4+11+14+21+...+(5n+(-1)^n

#### Yankel

##### Active member
Dear all

I am trying to prove by induction the following:

I checked it for n=1, it is valid. Then I assume it is correct for some k, and wish to prove it for k+1, got stuck with the algebra. Can you kindly assist ?

Thank you.

#### MarkFL

##### Pessimist Singularitarian
Staff member
I would state the induction hypothesis $$P_n$$:

$$\displaystyle \sum_{k=1}^{n}\left(5k+(-1)^k\right)=\frac{1}{2}\left(5n(n+1)+(-1)^n-1\right)$$

As the induction step, I would add $$\displaystyle 5(n+1)+(-1)^{n+1}$$ to both sides:

$$\displaystyle \sum_{k=1}^{n}\left(5k+(-1)^k\right)+5(n+1)+(-1)^{n+1}=\frac{1}{2}\left(5n(n+1)+(-1)^n-1\right)+5(n+1)+(-1)^{n+1}$$

Incorporate the new term:

$$\displaystyle \sum_{k=1}^{n+1}\left(5k+(-1)^k\right)=\frac{1}{2}\left(5n(n+1)+(-1)^n-1+2\cdot5(n+1)+2\cdot(-1)^{n+1}\right)$$

$$\displaystyle \sum_{k=1}^{n+1}\left(5k+(-1)^k\right)=\frac{1}{2}\left(5(n+1)(n+2)+(-1)^n(1+2(-1))-1\right)$$

$$\displaystyle \sum_{k=1}^{n+1}\left(5k+(-1)^k\right)=\frac{1}{2}\left(5(n+1)((n+1)+1)+(-1)^{n+1}-1\right)$$

We have derived $$P_{n+1}$$ from $$P_n$$ thereby completing the proof by induction.