One can compute the Betti numbers of a smooth complete intersection $X$ of multidegree $d=(d_1,\ldots,d_r)$ by induction on $r$. The case $r=0$ corresponds to $\mathbb{P}^n(\mathbb{C})$.
A smooth complete intersection of $X$ multidegree $(d_1,\ldots, d_r)$ is a section of the positive line bundle $\mathcal{O}(d_1)$ over a smooth complete intersection of $Y$ multidegree $(d_2,\ldots, d_r)$. By the Lefschetz hyperplane theorem we have $H^i(X,\mathbb{Q})\cong H^i(Y,\mathbb{Q})$ for $i\leq \dim Y-2=\dim X -1$. Using the induction hypothesis and the Poincar'e duality we can compute all $H^i(X,\mathbb{Q})$ apart from $H^{\dim X}(X,\mathbb{Q})$.
But we can also compute the Euler characteristic of $X$ as follows. The normal bundle of $X$ is $\mathcal{O}(d_1)\oplus\cdots\oplus\mathcal{O}(d_r)$, which extends to $\mathbb{P}^n$. The Chern class of $T\mathbb{P}^n$ is $(1+H)^{n+1}$ where $H$ is the class of the hyperplane section. So the Chern class of $TX$ is $$\frac{(1+H)^{n+1}}{\Pi_{i=1}^{r}(1+d_i H)}$$ restricted to $X$. So the Euler characteristic of $X$ is $d_1\cdots d_r$ times the coefficient of $H^{\dim X}$ in the above expression. Now we can compute the rank of $H^{\dim X}(X,\mathbb{Q})$.
Moreover, from the above and the universal coefficients formula it follows that $H^*(X,\mathbb{Z})$ is torsion free. Indeed, we may assume that $H_{<\dim X}(X,\mathbb{Z})$ and $H^{<\dim X}(X,\mathbb{Z})$ are torsion free by induction hypothesis and hence so are $H_{>\dim X}(X,\mathbb{Z})$ and $H_{>\dim X}(X,\mathbb{Z})$ by the Poincar'e duality; moreover $H^{\dim X}(X,\mathbb{Z})=Hom (H^{\dim X}(X,\mathbb{Z}),\mathbb{Z})$ is torsion free and so is $H_{\dim X}(X,\mathbb{Z})$ by the Poincar'e duality again.
