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Friday, March 4, 2016

Analysis Quiz 21: Multiple-Choice Test (Improve Logical And Common Sense Reasoning In Learning Mathematics)

Question 1: What can you conclude to the sum of the series based on the sequence listed as follows?
P=110100+1+110100+2+110100+2++110100+10

A. P>1.
B. P>10.
C. P>100.

Answer:

As long as the least in the given decreasing sequence is greater than 1, then all of the terms listed in the sequence are greater than 1. And the least term falls on the the last term, i.e. 110100+10=1.

Therefore, all of them are greater than 1. Since there are 10 terms in the series, then their sum must be greater than 1×10=10.

Question 2: You're asked to prove the RHS of the inequality below by rewriting 110 as a sum of some numbers, which of the following option would you adopt?

111<1100+1+1100+2++1100+10110

A. 110=1100+1100++110010terms

B. 110=150+125+125

C. 110=160+148+148+148+148

Answer:

Symmetry is what is crucial in determining the answer for this question. We have 10 terms in the series, it's necessary to write 110 as a sum of 10 terms.

Thus A is the answer.

Question 3: Are you able in using the hints provided in the preceding questions to prove the RHS of the inequality
111<1100+1+1100+2++1100+10<110?

A. Yes.
B. No.

Answer:

We want to prove the RHS of the given inequality above, i.e.

1100+1+1100+2++1100+10<110

When we've decided to rewrite 110 as a sum of 1100+1100++110010terms, we'll start to work on there:

1100+1+1100+2++1100+10<1100+1100++110010terms

It's not hard to note that each of the term on the left is less than 1100:

1100+1=1101<1100

1100+2=1102<1100

1100+3=1103<1100



1100+10=1110<1100

Therefore, we can conclude that

1100+1+1100+2++1100+10<10100<110

Question 4: What algebraic modification would you apply to the fraction 111 if you're to use the similar trick in proving the RHS of the inequality to prove the LHS of the inequality?

111<1100+1+1100+2++1100+10<110?

A. 111=155+155++1555terms

B. 111=188+188++1888terms

C. 111=1110+1110++111010terms

Answer:

Again, we have 10 terms in 1100+1+1100+2++1100+10, it's only appropriate to also create another 10 terms from 111, therefore we know C is the answer, and we need to verify that our choice works by working on proving it to be correct.

Note that

1110<1100+1=1101

1110<1100+2=1102



1110=1100+10=1110

Adding them all up yields the desired result:

111<1100+1+1100+2++1100+10

Question 5: Let P=1100+1+1100+2++1100+10 and that 111<P<110, determine the value for X,Y if X<1P<Y.

A. (X,Y)=(10,11)

B. (X,Y)=(100,110)

C. (X,Y)=(111,110)

D. (X,Y)=(1110,1100)

Answer:

This is a breeze if you've a strong mathematics foundation as:

If 111<P<110, then 10<1P<11.

That is, if you invert both sides of the inequality, (that is, flip fractions upside down on both sides), you have got to reverse the direction of the inequality sign.

Therefore A is the answer.

Question 6: Can you prove the following inequality effectively using the trick you used in the previous problems?

Let Q=120172+1+120172+2+120172+3++120172+2017

2017<1Q<2018

A. Yes.
B. No.

Answer:

You should be able to process the following procedures after you learned the tricks in the above questions:

120172+1+120172+2+120172+3++120172+2017 is a decreasing series with a total number of terms of 2017, and largest term is the first term, 120172+1 and the smallest term is the last term, 120172+2017.

If the smallest term, the last term (120172+2017=12018(2017))12018(2017), then all of the terms will be greater than 12018(2017).

We then can make the partial conclusion that Q=120172+1+120172+2+120172+3++120172+2017>12018(2017)×2017=12018.

By the same token, if the largest term, the first term in the series of Q, 120172+1  is less than 120172, which is true, then all of the remaining terms will be less than 120172.

We can then make the another half conclusion,
Q=120172+1+120172+2+120172+3++120172+2017<120172×2017=12017.

Therefore, we get:

12018<Q<12017

And so

2017<1Q<2018 (Q.E.D).


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