Sum of integers in table

nafiul9090 wrote:what if last two rows are omitted?? is there any uniform method or formula....

Solution:
Let us say that the first row is 1,2,3,4,5,6…n.
So the sum of all the above numbers is {n*(n+1)/2}.
Now, the sum of numbers in second row is -2*n(n+1)/2, in the third row is +3*n(n+1)/2…and so on.
If this extends till k where k is even, then the sum of last row is –k*n(n+1)/2.
If k is odd, then the sum of last row is +k*n(n+1)/2.
Hence the sum of all integers where k is even is {n*(n+1)/2}*{1-2+3-4…-k) = {n*(n+1)/2}*{-k/2}
= -k*n*(n+1)/4.
Also, the sum of all the integers where k is odd is {n*(n+1)/2}*{-(k-1)/2+k} = {n*(n+1)/2}*{(k+1)/2}.
If the last 2 rows are removed, then n= 7, k = 5 and is odd. So, the sum is {7*(7+1)/2}*{(5+1)/2} = 28*3 = 84