$E(n)\# \mathbb{CP}^2=2n\mathbb{CP}^2\# (10n-1)\overline{\mathbb{CP}^2}$

Question

How does one prove the following statement : $$E(n)\# \mathbb{CP}^2=2n\mathbb{CP}^2\# (10n-1)\overline{\mathbb{CP}^2}$$ Here $E(n)$ denotes the $n$-th elliptic surface formed by fiber summing $E(1)$ $n$ times. I tried to do induction on $n$, so for $n=1$, I saw that the definition of the elliptic surface $E(1)$ is $\mathbb{CP}^2\# 9\overline{\mathbb{CP}^2}$ so it holds in that case. But for $n=2$ I am getting $$E(2)\# \mathbb{CP}^2=3\mathbb{CP}^2\# 18\overline{\mathbb{CP}^2}$$ but the statement claims it should be $4\mathbb{CP}^2\# 19\overline{\mathbb{CP}^2}$. How does one show the numbers should match? Will this require Kirby diagrams? I'm not very expert in Kirby diagrams, so any help would be appreciated.