Match the following Column I to Column II

Answer

`(A) to (p, q); (B) to (p, r, t); (C) to (s, t)`

Explanation

For `lambda = 1, a = sum _(r=0)^(n) C_(r) = 2^(n)`
For `lambda = r, b = sum _(r=0)^(n) r C_(r) = sum _(r=0)^(n) r cdot n/r cdot ""^(n-1)C_(r-1)`
`= n sum _(r=o)^(n) ""^(n-1)C_(r-1) = n cdot 2^(n-1) `
and for `lambda = r^(2) , c = sum _(r=0)^(n) r^(2) C_(r) = sum _(r=0)^(n) r^(2) cdot n/r cdot ""^(n-1)C_(r-1)`
` = n sum _(r=0)^(n) rcdot ""^(n-1) C_(r) = nsum _(r=1)^(n) r cdot ""^(n-1)C_(r-1)`
` = n [sum _(r=0)^(n){(r-1) +1} ^(n-1) C_(r-1)]`
` = n [sum _(r=0) ^(n)(r-1)cdot ^(n-1) C_(r-1)+sum_(r=1)^(n)""^(n-1) C_(r-1)]`
` = n [sum _(r=0) ^(n)(r-1)cdot ((n-1))/((r-1))cdot""^(n-2) C_(r-2)+ 2^(n-1)]`
` = n [(n-1)sum _(r=0) ^(n) ""^(n-2) C_(r-2)+ 2^(n-1)]`
`= n [ (n-1) cdot 2^(n-2) + 2^(n-1) ] = n (n=1) 2^(n-2)`
`{:("For " n= 1, a = , b=1, c =1),("and for " n = 2, a=4, b=4, c=6):}}{:(a = b +c),(a + b = c +2):}`
(B) For `lambda = 1, a= sum_(r=0)^(n) (-1)^(r) cdot C_(r) = 0`
For `lambda = r`,
`b = sum_(r=0)^(n) (-1)^(r) .r. C_(r) = sum_(r=0)^(n) (-1)^(r) .r. (n)/(r) ""^(n-r)C_(r-1)`
` = n sum_(r=1)^(n) (-1)^(r). ""^(n-1)C_(r-1) = n(1-1)^(n-1) = 0 `
and for ` lambda = r^(2) ,c = sum _(r=0)^(n) (-1)^(r) .r^(2) . C_(r)`
`= sum_(r=0)^(n) (-1)^(r) .r^(2).(n)/(r) ""^(n-1)C_(r-1)`
` = n sum _(r=0)^(n) (-1)^(r) .r. ""^(n-1)C_(r-1)`
` = n sum_(r=0)^(n) (-1)^(r) {(r-1)+ 1}""^(n-1)C_(r-1)`
` = sum_(r=0)^(n) (-1)^(r) ""^(n-1)_(r-1_) + n sum_(r=0)^(n) (-1)^(r). ""^(n-1)C_(r-1)`
` = 0 + 0 = 0 `
` therefore a = b = c= 0 rArr a = b + c `
`rArr a^(3) + b^(3) + c^(3) = 3abc rArr a + b = 4b `
(C) For `lambda = 1,a = sum_(r=0)^(n) (C_(r))/((r +1)) = (1)/((n+1)) sum_(r=0)^(n) ((n +1)/(r +1)). ""^(n)C_(r)`
` = (1)/((n+1)) sum_(r=0)^(n) ""^(n+1)C_(r+1) = (1)/(n+1) (2^(n+1) -1)`
` = (2^(n+1)-1)/(n+1)`
For ` lambda = r , b = sum (r=0)^(n) (r.C_(r))/((r +1)) = sum_(r=0)^(n) ((r -1) + (1)/(r+1))C_(r)`
`sum_(r=0)^(n) r. C_(r) - sum_(r=0)^(n) C_(r) + sum_(r=0)^(n) (C_(r))/(r +1)`
` = n. 2^(n-1)- 2^(n) + (2^(n+1)-1)/(n+1)`
` = ((n^(2) - n + 2 ) 2^(n-1) -1)/((n+1))`
`{:(" for n " = 1, a = (3)/(2) , b = (1)/(2),c= (1)/(2)),(" and for "n = 2 , a = (7)/(3), b = (5)/(3), c= (7)/(3)):}}{:(a + b = 4b ),(b^(c-a) + (c - a)^(b)= 1):}`