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## Number Properties - Pattern Recognition for Sum of Cubes

Today, let’s take up a question in which pattern recognition will help us. Mind you, there are various ways of solving a question. Most questions we solve using pattern recognition can be solved using some other method. But pattern recognition is a method we can use in various cases. It is something that comes to our aid when we forget everything else. If you don’t know from where to start on a question, try to give some values to the variables. You might see a pattern. You may not ‘know’ something. Even then, you can ‘figure out’ the answer because GMAT is not a test of your knowledge; it is a test of your wits. It is a test of whether you can keep your cool when faced with the unknown and use whatever you know to solve the question.

Let’s look at a question now.

Question: What is the sum of the cubes of the first ten positive integers?

(A) 10^3

(B) 45^2

(C) 55^2

(D) 100^2

(E) 100^3

Solution:

You are obviously not expected to find the cubes and sum them.

There is a simple formula to find the sum of cubes of first n positive integers. It is given by [n(n+1)/2]^2
Put n = 10 in the formula. You get [10*11/2]^2 = 55^2.

Now, do we need to learn the formula?
No. Even if you didn’t know it, you should have tried to look at the pattern.

1^3 = 1

1^3 + 2^3 = 9

1^3 + 2^3 + 3^3 = 36

1^3 + 2^3 + 3^3 + 4^3 = 100

We see that all the sums are squares.

1^3 = 1 = 1^2

1^3 + 2^3 = 9 = 3^2

1^3 + 2^3 + 3^3 = 36 = 6^2

1^3 + 2^3 + 3^3 + 4^3 = 100 = 10^2

The point is – what are they squares of? What are 1, 3, 6 and 10? If you notice carefully, you will see that we obtain these numbers when we add the first n numbers.

1 = 1

1 + 2 = 3

1 + 2 + 3 = 6

1 + 2 + 3 + 4 = 10

This is pattern recognition. It might seem a little hard initially, but once you get used to it, you realize that every number you get is there for a reason. E.g., very rarely will you see 81 if it has nothing to do with it being 3^4.

You can also use another method in this question – of averaging.

We need to find this sum: 1^3 + 2^3 + 3^3 + … + 8^3 + 9^3 + 10^3

The numbers on the right 8^3, 9^3, 10^3 will be much larger than those on the left which are small. The average would lie not in the middle at 5^3 but on the right somewhere between 6^3 and 7^3. I would say around 300. Since the average represents the number that can replace every number in the list, the sum will be around 300*10 = 3000. This leads us to 55^2 since 55^2 = 3025 (it is very easy to find squares of numbers ending with 5. We will discuss it soon.)

The only hitch in using this approximation is the 45^2.

45^2 = 2025

To convince yourself that the average is around 300 and not around 200, notice that from 10^3 = 1000, you can make five 200s. From 9^3, you can make about four 200s (using some extra). So overall, you can make many more 200s than the required 10. Therefore, the average must be around 300 and not around 200.