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

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.