Show that : (^(4n)C_(2n))/(^(2n)C_n) = (1.3.5...(4n-1))/{1.3.5...(2n-1)}^2

Prove that , .^(2n)C_(n)=2^(n)(1.3.5...(2n-1))/(lfloorn)

If (1+x)^(n)=C_(0)+C_(1)+x+C_(2)x^(2)+...+C_(n) x^(n) Show that C_(1)^(2)+2*C_(2)^(2)+3*C_(3)^(2)....+n*C_(n)^(2)=((2n-1)!)/([(n-1)!]^(2))

Find the sum of the series .^(n)C_(0)+2.^(n)C_(1)x+3.^(n)C_(2)x^(2)+....+(n+1).^(n)C_(n)x^(n) and hence show that , .^(n)C_(0)+2.^(n)C_(1)x+3.^(n)C_(2)x^(2)+....+(n+1)^(n)C_(n)=(n+2)2^(n-1)

If (1+x)^(n)=C_(0)+C_(1)x+C_(2)x^(2)+...+C_(n)x^(n) "show that ", (1)/(1!(n-1)!)+(1)/(3!(n-3)!)+(1)/(5!(n-5)!)+....=(2^(n-1))/(n!)

Show that , (2n)! =2^(n).n![1.3.5…(2n-1)].

Prove that .^(2n)P_(n)={1.3.5.....(2n-1)}.2n

Prove that (2n!)/(n!)={1.3.5.....(2n-1)}2^n

Prove that .^(n)C_(1) + 2 xx .^(n)C_(2) + 3 xx .^(n)C_(3) + "…." + n xx .^(n)C_(n) = n2^(n-1) . Hence, prove that .^(n)C_(1).(.^(n)C_(2))^(2).(.^(n)C_(3))^(3)"......."(.^(n)C_(n))^(n) le ((2^(n))/(n+1))^(.^(n+1)C_(2)) AA n in N .

Prove that (.^(2n)C_0)^2-(.^(2n)C_1)^2+(.^(2n)C_2)^2-..+(.^(2n)C_(2n))^2 = (-1)^n.^(2n)C_n .

Show that the value of the middle term in the expansion of (x+(1)/(2x))^(2n)" is " (1.3.5.....(2n-1))/(n!)