InterviewSolution
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InterviewSolution
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For any positive integers m, n (with `n ge m`) If `({:(n),(m):}) = .^(n)C_(m)` Prove that `({:(n),(m):}) + ({:(n - 1),(m):}) + ({:(n - 2),(m):}) + … + ({:(m),(m):}) = ({:(n + 1),(m + 1):})` Prove that `({:(n),(m):}) + 2 ({:(n + 1),(m):}) + 3 ({:(n - 2),(m):}) + .... + (n - m + 1)` `({:(m),(m):}) = ({:(n + 2),(m + 2):})` |
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Answer» Let `S = ((n)/(m))+((n-1)/(m)) + ((n-2)/(m))+ ((n-2)/(m)) +.....+((m)/(m)) =((n+1)/(m+1)).....(i)` It is obvious that, `n ge m` Note : This question is based upon additive loop. Now ,`S =((m)/(m)) + ((m+1)/(m)) + ((m+2)/(m)) +.......+((n)/(m))` ` ={((m+1)/(m+1))+((m+1)/(m))}[because ((m)/(m)) = 1= ((m+1)/(m+1))]` `= ((m+2)/(m+1)) + ((m+2))/(m)) + ......+ ((n)/(m))" "[because""^(n)C_(r)+ ""^(n)C_(r+1) = ""^(n+1)C_(r+1)]` `=((m+2)/(m+1)) +......+((n)/(m))` `=.............` ` =((n)/(m+1))+((n)/(m)) = ((n+1)/(m+1))` which is ture ....(ii) Again, we have to prove that `((n)/(m))+2((n-1)/(m)) + 3((n-2)/(m)) +......+ (n-m+1)((m)/(m)) = ((m+2)/(m+2))` Let `S_(1) = ((n)/(m))+2((n-1)/(m)) +3((n-2)/(m)) +......+(n-m+1)((mm)/(m))` `{:(= ((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 Now, sum of the first row is `((n+1)/(m+1))` Sum of the second row is `((n)/(m+1))` Sum of the third row is `((n+1)/(m+1))`, .................... Sum of the last row is `((m)/(m)) = ((m+1)/(m+1))` 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))` [from Eq. (i) replacing n by n +1 and m by m + 1] |
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