for any positive integers m, n (with `n ge m`) <br> if `({:(n),(m):}) = .^(n)c\_(m)` prove that <br> `({:(n),(m):}) + ({:(n - 1),(m):}) + ({:(n - 2),(m):}) + … + ({:(m),(m):}) = ({:(n + 1),(m + 1):})` <br> prove that <br> `({:(n),(m):}) + 2 ({:(n + 1),(m):}) + 3 ({:(n - 2),(m):}) + .... + (n - m + 1)` <br> `({:(m),(m):}) = ({:(n + 2),(m + 2):})`
for any positive integers m, n (with `n ge m`) <br> if `({:(n),(m):}) = .^(n)c\_(m)` prove that <br> `({:(n),(m):}) + ({:(n - 1),(m):}) + ({:(n - 2),(m):}) + … + ({:(m),(m):}) = ({:(n + 1),(m + 1):})` <br> prove that <br> `({:(n),(m):}) + 2 ({:(n + 1),(m):}) + 3 ({:(n - 2),(m):}) + .... + (n - m + 1)` <br> `({:(m),(m):}) = ({:(n + 2),(m + 2):})`




solution: let `s = ((n)/(m))+((n-1)/(m)) + ((n-2)/(m))+ ((n-2)/(m)) +.....+((m)/(m)) =((n+1)/(m+1)).....(i)` <br> it is obvious that, `n ge m` <br> note : this question is based upon additive loop. <br> now ,`s =((m)/(m)) + ((m+1)/(m)) + ((m+2)/(m)) +.......+((n)/(m))` <br> ` ={((m+1)/(m+1))+((m+1)/(m))}[because ((m)/(m)) = 1= ((m+1)/(m+1))]` <br> `= ((m+2)/(m+1)) + ((m+2))/(m)) + ......+ ((n)/(m))" "[because""^(n)c\_(r)+ ""^(n)c\_(r+1) = ""^(n+1)c\_(r+1)]` <br> `=((m+2)/(m+1)) +......+((n)/(m))` <br> `=.............` <br> ` =((n)/(m+1))+((n)/(m)) = ((n+1)/(m+1))` which is ture ....(ii) <br> again, we have to prove that <br> `((n)/(m))+2((n-1)/(m)) + 3((n-2)/(m)) +......+ (n-m+1)((m)/(m)) = ((m+2)/(m+2))` <br> let `s\_(1) = ((n)/(m))+2((n-1)/(m)) +3((n-2)/(m)) +......+(n-m+1)((mm)/(m))` <br> `{:(= ((n)/(m)) + ((n-1)/(m)) + ((n-2)/(m)) +...+ ((m)/(m))), ( " "+ ((n-1)/(m)) + ((n-2)/(m)) +...+ ((m)/(m))), (" "+ ((n-2)/(m)) +...+((m)/(m)) ),(" "...) , (" "+ ((m)/(m))):}}n-m + 1` rows <br> now, sum of the first row is `((n+1)/(m+1))` <br> sum of the second row is `((n)/(m+1))` <br> sum of the third row is `((n+1)/(m+1))`, <br> .................... <br> sum of the last row is `((m)/(m)) = ((m+1)/(m+1))` <br> thus `s = ((n+1)/(m+1))+((n)/(m+1)) + ((n +1)/(m+1))+.......+ ((m+1)/(m+1)) = ((n+1+1)/(m+2)) = ((n+2)/(m+2))` <br> [from eq. (i) replacing n by n +1 and m by m + 1]