If `sum_(k=1)^(n+1) ^(n-k)C_r=x_y` then `x=nandy=r+1`

If sum_(k=1)^(n+1)hat imath(n-k)C_(r)=x_(y) then x=n and y=r+1

Prove that sum_(k=1)^(n-r ) ""^(n-k)C_(r )= ""^(n)C_( r+1) .

If {:("n-r"),( sum),("k=1"):} .^(n-k)C_(r)=.^(x)C_(y) then

If n in N, sum_(k=1)^(n)cos^(-1)(x_(k))=npi then the value of sum_(k=1)^(n)sin^(-1)(x_(k))=

If sum_(r=1)^(n)cos^(-1)x_(r)=0, then sum_(r=1)^(n)x_(r) equals to

If n in N, sum_(k=1)^(n)sin^(-1(x_(k))=(npi)/2 then the value of sum_(k=1)^(n)x_(k)=

If sum_(r=1)^(n)Cos^(-1)x_(r)=0," then "sum_(r=1)^(n)x_(r) equals to

If sum_(r=1)^(n) t_(r ) = sum_(k=1)^(n) sum_(j=1)^(k) sum_(i=1)^(j) 2 , then sum_(r=1)^(n) (1)/( t_(r )) equals :