What is the sum of (2^3 - 1)/1 + (3^3 - 1)/2 + (4^3 - 1)/3 + (5^3 - 1)

Question

What is the sum of  \frac{2^3 - 1}{1} + \frac{3^3 - 1}{2} + \frac{4^3 - 1}{3} + \frac{5^3 - 1}{4} + \frac{6^3 - 1}{5} + ... + \frac{20^3 - 1}{19} ?

A. 3,100
B. 3,099
C. 3,097
D. 3,044
E. 3,025

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GMATinsight · Accepted Answer

That's a really hard and lengthy problem to solve,.. But the game is all about the unit digit.

Please check the video explanation here Answer: Option C Formulas we need: Sum of n consecutive positive integers starting from 1, ∑n = 1+2+3+.....+n = (\frac{1}{2})*n*(n+1) Sum of n consecutive positive EVEN integers starting from 2, ∑2n = 2+4+6+8...+2n = n*(n+1) Sum of n consecutive positive ODD integers starting from 1, ∑(2n-1) = 1+3+5.....+(2n-1) = n^2 Sum of n positive squares starting from 1, ∑n^2 = 1^2+2^2+3^2+.....+n^2 = (\frac{1}{6})n*(n+1)(2n+1) Sum of n consecutive positive Cubes starting from 1, ∑n^3 = 1^3+2^3+3^3+.....+n^3 = ^2 The sum of cubes is not needed for GMAT but I mentioned it just because it belongs to the same family

https://www.youtube.com/watch?v=1Li6aCGKOUo

preetamsaha · Answer

T1= (2^3-1)/(2-1) 
 =(2-1)(2^2+2*1+1^1) /(2-1)    
 =(2^2+2*1+1^1)
similarly, T2= (3^3-1)/(3-1) 
 =(3-1)(3^2+3*1+1^1) /(3-1)
 =(3^2+3*1+1^1)
so, S= T1+T2+...+T19
 =   +   +  
 =   +   + 17 
 = 2870 + 210 + 17
 =3097
 correct answer C

sambitspm · Answer

See the attachment

dave13 · Answer

GMATinsight can you please list all formulas with their names .... like this is formula for finding this ...... etc ... that you used in your solution

dave13 · Answer

during actual exam its unrealistic i think to use standard way to solve such question, at least for me, so i thought since there are 19 terms and the 19th term is odd, then its highly probable that answer should be odd number , so i eliminated A and D lol now left with B, C and E.
E eliminated since unit digit 5 didnt look fit at all lol so i was left with B and C , here i got stuck and chose B. i wonder if there was a trick to choose C in my case

GMATinsight · Answer

dave13 Here are the formulas we need to know Formulas: Sum of n consecutive positive integers starting from 1, ∑n = 1+2+3+.....+n = (\frac{1}{2})*n*(n+1) Sum of n consecutive positive EVEN integers starting from 2, ∑2n = 2+4+6+8...+2n = n*(n+1) Sum of n consecutive positive ODD integers starting from 1, ∑(2n-1) = 1+3+5.....+(2n-1) = n^2 Sum of n positive squares starting from 1, ∑n^2 = 1^2+2^2+3^2+.....+n^2 = (\frac{1}{6})n*(n+1)(2n+1) Sum of n consecutive positive Cubes starting from 1, ∑n^3 = 1^3+2^3+3^3+.....+n^3 = ^2 The sum of cubes is not needed for GMAT but I mentioned it just because it belongs to the same family

Onyedelmagnifico · Answer

Note that   =1+2+4
Similarly, 
  = 1+20+400.
This implies that the above series can be written as 
1+2+4 + 1+3+9+ 1+4+16+ ...+ 1+20+400 
This can be written as 
{1+2+3+4+...+20} + {1+4+9+16+...+400}+17=
210+ 2870 +17= 3097 {option C}

Posted from my mobile device

ScottTargetTestPrep · Answer

Notice that all the terms are in the form of (n^3 - 1)/(n - 1) for n = 2, 3, …, 20.

Since n^3 - 1 = (n - 1)(n^2 + n + 1), then each term can be simplified as n^2 + n + 1 and we have to add these terms from n = 2 to 20. In other words, we have to sum the n^2 terms from n = 2 to 20, the n terms from n = 2 to 20, and also 19 terms of 1.

Recall that 1^2 + 2^2 + 3^2 + … + k^2 =  /6 and 1 + 2 + 3 + … + k = k(k + 1)/2, so 2^2 + 3^2 + … + k^2 =  /6 - 1 and 2 + 3 + … + k = k(k + 1)/2 - 1.

Therefore, the sum of the n^2 terms from n = 2 to 20 is:

2^2 + 3^2 + … + 20^2 =  /6 - 1 =  /6 - 1 = 10(7)(41) - 1 = 2,869

The sum of the n terms from n = 2 to 20 is:

2 + 3 + … + 20 = 20(20 + 1)/2 - 1 = 20(21)/2 - 1 = 10(21) - 1 = 209

Last but the least, we also have to add 19 terms of 1, which is 19 x 1 = 19.

Therefore, the sum of the given expression is 2,869 + 209 + 19 = 3,097.

Answer: C

Kinshook · Answer

Asked: What is the sum of  \frac{2^3 - 1}{1} + \frac{3^3 - 1}{2} + \frac{4^3 - 1}{3} + \frac{5^3 - 1}{4} + \frac{6^3 - 1}{5} + ... + \frac{20^3 - 1}{19} ?

t_n = \frac{n^3 - 1}{(n-1)} = (n^2 + n + 1) = n(n+1) +1 = \frac{n(n+1)(n+2)}{3} - \frac{(n-1)n(n+1)}{3} + 1

\frac{2^3 - 1}{1} + \frac{3^3 - 1}{2} + \frac{4^3 - 1}{3} + \frac{5^3 - 1}{4} + \frac{6^3 - 1}{5} + ... + \frac{20^3 - 1}{19} = \frac{20*21*22}{3} - \frac{1*2*3}{3} + 19 = 3080 - 2 + 19 = 3097

IMO C

onlymalapink · Answer

non of the solutions make intuitive sense to me. I cant see the logic in the process at all.

onlymalapink · Answer

why are you adding one in both sequences what's the logic you are already adding 17

What is the sum of (2^3 - 1)/1 + (3^3 - 1)/2 + (4^3 - 1)/3 + (5^3 - 1)

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What is the sum of (2^3 - 1)/1 + (3^3 - 1)/2 + (4^3 - 1)/3 + (5^3 - 1)

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