Question

find the sum of all the numbers from 1 to 8027 whose cube roots are natural​

Answer 1

Answer:

210

Step-by-step explanation:

Numbers from 1 to 8027 whose cube roots are natural : n

$\sqrt[3]{1}=1$

$20<\sqrt[3]{8027} <21$

∴n=1, 2, ..., 20

Sum : $21\times \frac{20}{2} =210$

Answer 2

Given:

The cube roots lie between the cube root of 1 and cube root of 8027

To Find:

The sum of all the numbers from 1 to 8027 whose cube roots are natural.

Solution:

Given,

Cube roots lie between the cube root of 1 and cube root of 8027.

First,

​We must first find the cube root of 1 and 8027:

$\sqrt[3]{1} = 1$

$\sqrt[3]{8027} = 20.02247473$

          = 20 ( Rounded to nearest whole number )

∴ there are 20 numbers with natural cube roots between 1 and 8027

    1, 2, 3, 4, 5, ............., 19, 20

Now,

• To find the sum of all the numbers, we can make use of the formula of "Sum of 'n' numbers in an Arithmetic Progression (A.P)"

S = $\frac{n}{2} [ 2a + ( n - 1 )] * d$

n= number of terms,    a= first term in the sequence,  

d= common difference between two terms

In this case,

n = 20 , a= 1 , d= 1

Put these value in the formula

S:

      $= \frac{20}{2} [2(1) + (20 - 1)] * 1\\\\= 10 ( 2 + 20 -1 )\\ = 10 ( 21 )\\= 210$

Hence, the sum of all the numbers from 1 to 8027 whose cube roots are natural​ is 210