Q:
C++ program to find Sum of cubes of first N Even numbers
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C++ programming
A simple solution is given below...
Example 1:
Output
The efficient approach is discussed below:
The sum of cubes of first n natural numbers is given by = (n*(n+1) / 2)^2 Sum of cubes of first n natural numbers can be written as... = 2^3 + 4^3 + .... + (2n)^3 Now take out common term i.e 2^3 = 2^3 * (1^3 + 2^3 + .... + n^3) = 2^3* (n*(n+1) / 2)^2 = 8 * ((n^2)(n+1)^2)/4 = 2 * n^2(n+1)^2Now we can apply this formula directly to find the sum of cubes of first n even numbers.
Example 2:
Output