Key Concepts: Combination and Counting, Head-Tail Shrink Method, Mid-Term Shrink Method, Sectional Shrink Method, Elimination Shrink Method

Hello young mathematicians! Today, we are going to explore a fascinating math problem that involves a series of fractions. Don’t worry, we will break it down step by step, and by the end of this article, you will be able to solve similar problems with ease.

Let’s take a look at the problem:

What is the integer part of $(1-\frac{1}{2}+\frac{1}{4}-\frac{1}{6}+\frac{1}{8}-\frac{1}{10}+…+\frac{1}{96}-\frac{1}{98}+\frac{1}{100}) \times 10$?

This problem might seem complicated at first glance, but with the right approach, it becomes a piece of cake. We will use the Elimination Shrink Method, which is perfect for situations where we have both addition and subtraction.

Let’s imagine that we remove the tail of the original sequence, everything from $+\frac{1}{8} -\frac{1}{10} + …+\frac{1}{96}-\frac{1}{98}+\frac{1}{100}$.

Now, is the part we removed greater than or less than zero?

$\frac{1}{8}-\frac{1}{10} >0$, $\frac{1}{12}-\frac{1}{14} >0$, …, $\frac{1}{96}-\frac{1}{98} >0$, $\frac{1}{100}>0$.

So, the part we removed is greater than zero. Therefore, $(1-\frac{1}{2}+\frac{1}{4}-\frac{1}{6}+\frac{1}{8}-\frac{1}{10}+…+\frac{1}{96}-\frac{1}{98}+\frac{1}{100}) \times 10 > 1-\frac{1}{2}+\frac{1}{4}-\frac{1}{6}$.

Next, let’s imagine that we remove the tail from a subtraction term, everything from $-\frac{1}{6} +\frac{1}{8} -\frac{1}{10} + …+\frac{1}{96}-\frac{1}{98}+\frac{1}{100}$.

Is the part we removed greater than or less than zero?

$-\frac{1}{6} +\frac{1}{8} <0$, $-\frac{1}{10} +\frac{1}{12} <0$, …, $-\frac{1}{98} +\frac{1}{100}<0$.

So, the part we removed is less than zero. Therefore, $(1-\frac{1}{2}+\frac{1}{4}-\frac{1}{6}+\frac{1}{8}-\frac{1}{10}+…+\frac{1}{96}-\frac{1}{98}+\frac{1}{100}) \times 10 < 1-\frac{1}{2}+\frac{1}{4}-\frac{1}{6}$.

So, we can see that every time we perform an addition, the result is slightly greater than the original value. Every time we perform a subtraction, the result is slightly less than the original value. When we calculate to the point where the tenth place does not change, the integer part of the result multiplied by 10 does not change.

Let’s visualize this:

$1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{6} + \frac{1}{8} - \frac{1}{10} + \frac{1}{12} - …$

$1↑ 0.5↓ 0.75↑ 0.583↓ 0.708↑ 0.608↓ 0.691↑$

So, $0.691 \times 10 >$ original sequence $> 0.608 \times 10$

Or, $6.91 >$ original sequence $> 6.08$

Therefore, the integer part of the original sequence is 6.

And there you have it! A seemingly complex problem solved with a simple and fun approach. Keep practicing, and soon you will be able to solve such problems in a jiffy. Happy learning!