What is the value of A=1/5^2+2/5^3+...+11/5^12?

Solution.

\(\begin{split}
A&=&\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{10}{5^{11}}+\frac{11}{5^{12}}\\
5A&=\frac{1}{5}+&\frac{2}{5^2}+\frac{3}{5^3}+\frac{4}{5^4}+...+\frac{11}{5^{11}}\\
5A-A&=\frac{1}{5}+&\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}+...+\frac{1}{5^{11}}-\frac{11}{5^{12}}\\
4A&=B-&\frac{11}{5^{12}}
\end{split}\)

\(\begin{split}
B&=&\frac{1}{5}+&\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}+...+\frac{1}{5^{10}}+\frac{1}{5^{11}}\\
5B&=1+&\frac{1}{5}+&\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}+...+\frac{1}{5^{10}}\\
4B&=1-&\frac{1}{5^{11}}\\
\end{split}\)
\(\begin{split}
4B&=\frac{5^{11}-1}{5^{11}}\\
B&=\frac{5^{11}-1}{4 \times 5^{11}}
\end{split}\)

Hence we have
\(\begin{split}
4A&=B-\frac{11}{5^{12}}=\frac{5^{11}-1}{4 \times 5^{11}}-\frac{11}{5^{12}}\\
4A&=\frac{5(5^{11}-1)-44}{4 \times 5^{12}}\\
4A&=\frac{5^{12}-5-44}{4 \times 5^{12}}\\
4A&=\frac{5^{12}-49}{4 \times 5^{12}}\\
A&=\frac{5^{12}-49}{16 \times 5^{12}}\\
A&=\frac{1}{16} \times \frac{5^{12}-49}{5^{12}}\\
A&=\frac{1}{16}\bigg( 1- \frac{49}{5^{12}} \bigg )\\
\end{split}\)

Since \(1- \frac{49}{5^{12}} < 1 \implies A < \frac{1}{16}\).

It's clear that the answer is C.

Also \(2 \times 49 < 5^{12} \implies \frac{49}{5^{12}} < \frac{1}{2} \implies 1- \frac{49}{5^{12}} > 1- \frac{1}{2}=\frac{1}{2}\)

Hence \(A>\frac{1}{16} \times \frac{1}{2}=\frac{1}{32}\).

This type of question is quite hard so It will probably not be shown on the real GMAT. However, the solution is long due to the fact that I want to exlain it in detail. In fact, I solve it under 2 mins. This question is just for enjoy solving math, so dont worry if you cant solve it.

Sure - first note the 4/63 becomes 4/64 so that you can simplify by 2. 4/63 does not become 2/64

4/63 -> 4/64 -> reduce to 2/32 -> reduce to 1/16

Does that clarify your question? I think you just misread the last part

First we will solve by approximation method. But be careful while selecting the range of numbers. i did mistake by selecting Between \(\frac{1}{16}\) and \(\frac{1}{8}\) instead of Between \(\frac{1}{32}\) and \(\frac{1}{16}\)

I figures out that the answer is near about 1/18 and 1/16 was a good approximation so I took the value between 1/16 and 1/8 but it should have been between 1/32 and 1/16

Now lets start the approximation..
Since \(\frac{1}{5^2}) = 1/25 . So the value must be near 1/25. But we will go further to check the nearest value.
\(\frac{2}{5^3}) = 2/125

So 1/25 + 2/125 = 7/125 > 1/18. Now here the confusion continues that if its between 1/16 and 1/8 or between 1/16 and 1/32.

To check we will surely have to go 1 step further, though the value of further terms is decreasing but it may cross the range of 1/16 and 1/32

1/25 + 2/125 + 3/625 = (25+10+3)/625 = 38/625 = 1/16.47

So if we really go by this method, we can't approximate the value... This question needs to be solved only through the long process. The options choice is not that easy to make a correct approximation... If the question would have been in GMAT format the options would not have been so close. There must have been wide gap between the options.

You have to tell how and what you got so that the your equation and the equation here can be compared.

Hi Bunuel any suggestion on how to solve this? the GP is for sure good, but looks to me to be too long for a 2 min problem

The way you solved looks great, but honestly for me hardly approachable on the test day due to time constraint, therefore I wanted to check if someone else had any other ways to approach it.

Thanks anyway clarification and the solution

Multiply both side by 1/5 and then subtract

4A/5 = 1/5^2 + 1/5^3.....

I considered it as infinite GP as it will give max value for A

4A/5 = (1/25 ) / (1 - 1/ 5 ) sum of infinite gp = a / 1- r

A=1/16 max value

1/25 falls into the range between 1/32 and 1/16. the rest are too small
Hence, the answer is C

Hi,

at first I was intimidated by the algebric computation, then I thought "No way! the GMAT won't give me something like that!"

A=1/25+1/72+(1/208 + ...) =(72+25)/(72*25)=1/18 which is between 1/32 and 1/16

1/208 is about 0.005 so, since the range is pretty broad 1/32=0.03 --> 1/16=(1/8=0.125)/2=0.0625, I rounded all the other terms to zero.

This way of reasoning brought me to C.

Hi - I solved in a little over a minute using the following method:

you have (1 / 5^2) + (2 / 5^3) + (3 / 5^4) ....whatever

First notice that the numbers become really small the further you go out so they don't really impact the fraction, just know that if the answer above ='s (as an example) 1/2 that the answer is >1/2 because it's 1/2 + whatever

Continuing, make the fractions all equal - so you have: (25 / 5^4) + (10 / 5^4) + (3 / 5^4)

Then add them to get: (38 / 5^4)

Next we need to estimate, so just assume 38 = 40 (so that it's divisible by one 5), to give you: 8 / 5(5^3)

Then reduce further again, so assume 125 = 126 so you can divide by 2 => 4 / 63
...again assume 63 = 64, so you have 2/64 => ~~1/32

finally, you have ~1/32 + some stuff (the "whatever" above) so it will be between 1/32 -> 1/16

Can you please explain how 4/63 became 2/64?

Hi,

This is how I solved this question and took me about 30 seconds total. We can use estimation here. If we take the first term and add it to me the very last term, we can arrive at the answer quite easily.

(1/5^2)+(11/5^12)= We don't even have to solve the entire thing. If we solve just the first part, we get 1/25 as our answer. That means the overall answer has to be in between 1/25 and (11/5^12). Hence, C

Please give me a "kudos" if you found my approach helpful!

Thanks!

You can solve this easily by taking 1/25 outside the bracket.Please note the answer choices are in range and not exact answers.The remaining terms inside the bracket would be(1+2/5+3/25).Take the first 3 terms only as anything beyond it is very minimal adding no value. You will end up in an intermediate solution of 1/25*(1.5) which makes you select option C as the right answer.

Are you sure if we subtract equation 1 from 2, we will get that GP series? Because when I try to do so I get different series than yours.

Hi broall, can't we simply re-write this as the sum from i=1 to i=11 of (n)/(5)^(n+1), and then note that the sum of the infinite geometric series written in this fashion would be a1/(1-r), so we could get (1/25)/(1-1/5), which is equal to 1/30, which is then bounded above by 1/16 and 1/30?

Just trying to minimize the calculation, let me know if this is a viable method for such problems. It does involve some Calculus 2 knowledge, specifically sums of infinite geometric series, but greatly simplifies the calculation.

A=1/5^2+...
A/5= 1/5^3+2/5^4+....11/5^13

4A/5=1/5^2+...1/5^12+11/5^13
=(1/5^2)(1-(1/5)^10)/(4/5) +11/5^13
=1/16(1-0.2^10)+11/5^13

slightly less than 1/16