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i have to show that:
1)[tex](-1)^n\int_{-1}^{1}(x^2-1)^ndx=2^{2n+1}(n!)^2/(2n+1)![/tex]
2) [tex]\binom{n}{k}=[(n+1)\int_{0}^{1}x^k(1-x)^{n-k}dx]^{-1}[/tex]
for the first part i thought to use Newton's binomial, i.e:
[tex](1-x^2)^n=\sum_{k=0}^{n}\binom{n}{k}(-x^2)^k[/tex]
but it didn't get me far.
for the second part i don't have a clue, i don't think you can integrate the integral by parts or substitution can you?
1)[tex](-1)^n\int_{-1}^{1}(x^2-1)^ndx=2^{2n+1}(n!)^2/(2n+1)![/tex]
2) [tex]\binom{n}{k}=[(n+1)\int_{0}^{1}x^k(1-x)^{n-k}dx]^{-1}[/tex]
for the first part i thought to use Newton's binomial, i.e:
[tex](1-x^2)^n=\sum_{k=0}^{n}\binom{n}{k}(-x^2)^k[/tex]
but it didn't get me far.
for the second part i don't have a clue, i don't think you can integrate the integral by parts or substitution can you?