find the sum of all the numbers of the form n^3which lie between 100 & 10000.?

Answer 1

100
cube root 100 < n < cube root 10000

4.64 < n < 21.5

So n goes from 5 to 21.

You have to add up 5^3 + 6^3 + 7^3 + ... + 21^3. That's a lot!

I get 53,261.

Answer 2

This project isn't as complicated because it first looks. that is not any longer something better than the sum of an mathematics progression. the first form between one hundred and ten thousand it truly is divisible by 3 is 102. so it truly is the first time period in our progression. The very last form in this determination is 9999. we may be able to then use right here formula to locate the fashion of words contained in the progression: A = a + (n-a million) d, the position A is the perfect time period, a the first time period, n the fashion of words and d the version, which subsequently is 3. 9999 = 102 + (n-a million) 3 9897 = (n-a million) 3 (9897)/3 = (n-a million) 3299 = (n-a million) 3300 = n The formula for the sum of an mathematics progression is: S = n[a + (n-a million) d/ 2]. we've each and every of the advice we opt to locate the sum now, so all we ought to do is plug them straight away into the formula. S = 3300[102 + (3299) 3/2] = 336600 + 16330050 = 16,666,650. a touch large sum certainly. on the different hand, if one needed to be facetious, you could nonetheless say that the sum of the numbers divisible by 3 which lie between one hundred and ten thousand is 0 because, as your question is truly posed, there are not any numbers.

Answer 3

In other words you want to find the sum of the series 5^3+6^3+...21^3. Use the formula 1^3+2^3+...n^3=(n^4+n^2+2n^3) / 4 to find sum of the series 1^3+2^3+...21^3 and subtract it from the series 1^3+2^3+3^3+4^3.

You should be getting 53,261 if you do the calculations correctly.

Answer 4

There's actually a closed form solution to the question. You want to sum,

5^3 + 6^3 + ... + 21^3.

But 1^3 + 2^3 + ... + n^3 =
((n+1) | 2) ^ 2 =
(n*(n+1)/2)^2 =
n^2 * (n+1)^2 / 4

So the answer is,

21*21*22*22/4 - 5*5*4*4/4 = 53361 - 100 = 53261.

Beats having to add 17 cubes together!!!

Answer 5

n^3 > 100 means n > 4
n^3 < 10000 means n < = 21
we have to find sum of cubes of 5 to 21

= sum of cubes of 1 to 21 - sum of cubes of 1 to 4
= ((22*21)/2)^2-((4*5)/2)^2
= (21*11)^2 - 10^2
= 231 ^2 - 100
= 53361- 100 = 53261

Answer 6

Start by finding what values of n are applicable. By trial and error you can see 4^3=64 is too low. 5^3=125 so 5 is the first n. 21^3=9261, 22^3=10648. That leaves 17 values for n, start punchin buttons.

Answer 7

21
∑ X³n = 53261
n = 5

Answer 8

summation of cubes formula from 1 to n is (n(n+1)/2)^2.

5^3=125, this is the first term in the sum.
21^2=9261, this is the last term in the sum.

the sum of cubes up to n = 21 is 53361, plugging into our summation formula n=21

subtract the sum of cubes up to n=4 because we want n = [5, 21]

the sum of cubes up to n=4 is 100, plugging into our summation formula n =4

S=53361-100=53261.

Answer 9

125+216+343+512+729+1000+1331+1728+2197+2744+3375+4096+4913+5832+6859+8000+9261=53261
the answer is 53261

Answer 10

n=5 to 21.
1^3 + 2^3 + 3^3 + ....+ n^3 = (1 + 2 + 3 + ...+n)^2
use this formula for n = 21 then just subtract off
the values of n^3 for n = 1,2,3,4