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Question
Prove that `"C"_0^2 + "C"_1^2 + "C"_2^2 + ... + "C"_"n"^2 = (2"n"!)/("n"!)^2`
Solution
We know `"C"_0 + "C"_1 + "C"_2 + ... + "C"_"n"` = 2n
And `"C"_0"C"_"r" + ""_1"C"_("r" + 1) + "C"_2"C"_("r" + 2) + ......... + "C"_("n" - "r")"C"_"n" = ""^(2"n")"C"_("n - r")`
Taking r = 0 we get
`"C"_0"C"_0 + "C"1"C"_1 + "C"_2"C"2 + ......... + "C"_"n""C"-"n" = ""^(2"n")"C"_"n"`
(i.e.) `"C"_0^2 + "C"_1^2 + "C"_2^2 + ......... + "C"_"n"^2`
= `""^(2"n")"C"_"n"`
= `(2"n"!)/("n"!(2"n" - "n")!)`
= `(2"n"!)/("n"!"n"!)`
= `(2"n"!)/("n"!)^2`
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