Sum of integers in table

[cmr209]

I've never encountered a question like this before.




What is the sum of the integers in the table above?

A 28

B 112

C 336

D 448

E 784

Please help.

[Morgoth]

There are many ways of solving this question, I'll show you 1

if you look at the rows closely, every row is a table of a number starting from 1 to 7


1+2+3+4+5+6+7 = 28

28 + -2(28) + 3(28) -4(28) + 5 (28) -6(28) + 7(28)

= 28 ( 1-2+3-4+5-6+7) = 4*28 = 112

Hope this helps.

[cmr209]

Yes! This does help, thank you!
I had never seen a problem like this.

[nafiul9090]

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

[Anurag@Gurome]

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

[GMATGuruNY]

I've never encountered a question like this before.




What is the sum of the integers in the table above?

A 28

B 112

C 336

D 448

E 784

Please help.

Look for a pattern.
The table shows the results of multiplying every integer in the top row by every integer in the leftmost column.

Thus, we need to determine the sum of all these products:
(Top row)*(Leftmost column) = (1+2+3+4+5+6+7)(1-2+3-4+5-6+7) = 28*4 = 112.

The correct answer is B.

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

Use the same reasoning as above, omitting the bottom 2 rows:
(Top row)*(Leftmost column) = (1+2+3+4+5+6+7)(1-2+3-4+5) = 28*3 = 84.