Closed form of :$ (n+1)^4y_{n+1}-n^4y_{n}=2n+1$

Question

I want to compute limit of $y_n$ for $ (n+1)^4y_{n+1}-n^4y_{n}=2n+1$ but i have accrossed a problem to get it's closed form i know only that the intial condition could be deduced by taking $n=0$ to get $y_{1}=1$, then my question is how i can get Closed form of :$ (n+1)^4y_{n+1}-n^4y_{n}=2n+1$ for the computation of limit ?

Answer 1

Hint: Just sum the equations $(n+1)^4y_{n+1}-n^4y_{n}=2n+1$ for $n=1,\dots, N-1$. There's lot of cancellation on the LHS and a familiar sum on the RHS.

Indeed, we get $N^4y_{N}-y_{1}=\sum_{n=1}^{N-1}(2n+1)=N^2-1$

Answer 2

Considering $$(n+1)^4y_{n+1}-n^4y_{n}=2n+1$$ define $z_n=n^4 y_n$ to get $$z_{n+1}-z_n=2n+1\implies z_n=n^2+C\implies y_n=\frac{n^2+C}{n^4}$$