To require the sum of this series, we first analyze this series.
The series is: 2,6,12,20,30...
For n = 6 Sum = 112 On analysis, (1+1),(2+4),(3+9),(4+16)... (1+1<sup>2</sup>), (2+2<sup>2</sup>), (3+3<sup>2</sup>), (4+4<sup>2</sup>), can be divided into two series i.e. s1:1,2,3,4,5… andS2: 1<sup>2</sup>,<sup>2</sup>,3<sup>2</sup>,....
Use mathematical formulas to find the sum of the first and second
Sum1 = 1+2+3+4… , sum1 = n*(n+1)/2 Sum2 = 12+2<sup>2</sup>+3<sup>2</sup>+4<sup>2</sup>… , sum1 = n*(n+1)*(2*n +1)/6
#include <stdio.h> int main() { int n = 3; int sum = ((n*(n+1))/2)+((n*(n+1)*(2*n+1))/6); printf("the sum series till %d is %d", n,sum); return 0; }
The sum of series till 3 is 20
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