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Ta có:
$\sum_{i=0}^{n}(C_n^i)^2=C_{2n}^n$ .
Thật vậy:
Xét biểu thức: $(x+1)^{2n}=\sum_{i=0}^{2n}a_ix^i$.
Ta có: $a_n=C_{2n}^n$.
Lại có: $(x+1)^{2n}=\Big(\sum_{i=0}^nC_n^ix^i\Big)^2$, nên: $a_n=\sum_{i=0}^nC_n^iC_n^{n-i}= \sum_{i=0}^{n}(C_n^i)^2$
Từ đó suy ra: $\sum_{i=0}^{n}(C_n^i)^2=C_{2n}^n$.
Quay lại bài toán:
$\sum_{i=0}^n\Big(\frac{C_n^i}{i+1}\Big)^2=\sum_{i=0}^n\Big(\frac{n!}{(i+1)!(n-i)!}\Big)^2$
$=\sum_{i=0}^n\Big(\frac{1}{n+1}C_{n+1}^i\Big)^2$
$=\frac{1}{(n+1)^2}\sum_{i=0}^n(C_{n+1}^i)^2=\frac{C_{2n+2}^{n+1}-1}{(n+1)^2}$
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