A table of numbers has n rows and n columns, where n is an odd integer

I created a table using n=5 (smallest case). Since we have negative/positive changes between rows, the sums will cancel each other out on the even rows (eg. rows 1 + 2 cancel, 3 + 4 cancel, etc). In the case of n=5, all we are left with is row #5. The entries in row 5 (left to right), n=5 is 1, 2, 4, 8, 16. Summing this gives you 31. Plugging n=5 into the answer choices, D gives you the correct answer.

Answer is D.

Answer = D

Refer diagram below:(5*5 matrix)



Only numbers in yellow remain

Sum = 1+2+4+8+16 = 31

\(2^5 - 1 = 31\)

\(Answer = 2^n - 1\)

Official Solution:

A table of numbers has \(n\) rows and \(n\) columns, where \(n\) is an odd integer greater than 4. Each entry in the table is subject to the following rules:

(I) Any entry is the negative of the entry directly above it, if there is an entry directly above it.

(II) Any entry is double the entry directly to its left, if there is an entry directly to its left.

If the entry in the upper left corner (first row and first column) is 1, what is the sum of all the entries in the table?

A. \(0\)
B. \(n^2 - 1\)
C. \(n^2 + 1\)
D. \(2^n - 1\)
E. \(2^n + 1\)

The best way to approach this problem is to generate an actual table according to the rules. Let's start with the most concrete information we know: the entry in the upper left corner is 1. Now let's build the table to the right and then down.

According to the second rule, entries double as we go to the right, so the first row looks like this:
\(\text{1, 2, 4, 8, 16 ...}\)

According to the first rule, entries "flipflop" in sign (+/-) as we go down the table. So the first five rows and columns look like this:
\(\text{1, 2, 4, 8, 16}\)
\(\text{-1, -2, -4, -8, -16}\)
\(\text{1, 2, 4, 8, 16}\)
\(\text{-1, -2, -4, -8, -16}\)
\(\text{1, 2, 4, 8, 16}\)

Since \(n\) is an odd integer greater than 4, we can choose \(n = 5\) and stop here.

The process of adding up all the entries benefits from cancellation. The first four rows cancel each other out completely. All that is left is the last row, which sums up as follows: \(1 + 2 + 4 + 8 + 16 = 31\). Plugging \(n = 5\) into the answer choices, we get the following:

(A) \(0\) INCORRECT

(B) \(n^2 - 1 = 52 - 1 = 24\) INCORRECT

(C) \(n^2 + 1 = 52 + 1 = 26\) INCORRECT

(D) \(2^n - 1 = 25 - 1 = 31\) CORRECT

(E) \(2^n + 1 = 25 + 1 = 33\) INCORRECT

We can prove that \(2^n - 1\) is the general formula for the sum of 1, 2, 4, etc. up to \(2^{n - 1}\) (which will be the last entry in the last row), but there is no need to do so.

Answer: D.

2^n means 2+2+(2+2)+(2+2+2+2)+(2+2+2+2+2+2+2+2)+...
since we double each entry, we go like this: 1+1*2+1*2*2+1*2*2*2 which is basically the same as above but with 1 in the beginning instead of 2. so if we calculate 2^n and remove the difference between 1 and 2 => -1,
we get: 2^n-1 (D)

Given that n is odd.

The first entry is 1. The remaining entries in the 1st row will be 1,2,4,8,16....

The entry in the 2nd row will be the negative of the entries in the 1st row.

Similarly, the entries in the 3rd row will negate those in the 2nd row.

Therefore, if there were even number of rows, the entries would be equally negated. Then the sum would have been 0.

Since there will be one row left (n is odd), the sum will be equal to the entry in the first row.

It will be equal to the sum of the powers of 2, which is given by (2^n) - 1.

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