Friday, 23 September 2016

Series Program : [(1^1)/1] + [(2^2)/2] + [(3^3)/3] + [(4^4)/4] + [(5^5)/5] + ... + [(n^n)/n]

[(1^1)/1] + [(2^2)/2] + [(3^3)/3] + [(4^4)/4] + [(5^5)/5] + ... + [(n^n)/n]
long power(int a, int b)
{
long i, p=1;
for(i=1;i<=b;i++)
{
p=p*a;
}
return p;
}

int main()
{
long i,n;
double sum=0;
n=5;
for(i=1;i<=n;i++)
{
sum=sum+(power(i,i)/i);
}
printf("Sum: %lf",sum);
return 0;
}

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