1.

If (1 + x)^(n) = C_(0) + C_(1)x + C_(2)x + ….. + C_(n) x^(n) then show that the sum of the products of the C_(i) is taken two at a time represented by sum sum C_(i) C_(j) is equal to 0 le I lt j n 2^(2n - 1) - ((2n!))/(2(n!)^(2)).

Answer»

Solution :We know that ,
`2SUM _(0) sum_(leile j lt n)sum C_i C_j=sum_(i=0)^(n)sum_(j=0)^(n)C_(i=0)^(n)sum_(j=0)^(n)C_i C_J- sum_(i=0)^(n)sum_(j=0)^(n)C_i C_j =sum_(i=0)^(n)C_i sum_(j=0)^(n)C_(j) -sum_(i=0)^(n)C_(i)^(2)`
`=2^n 2^n -(.^2n C_n)=2^2n -.^2n C_n`
`therefore sum sum_(0 LE i n j le n) C_iC_j=(2^2n -.^2n C_N)/(2)=2^(2n-1)-((2n)!)/(2(n!))`.


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