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void
sum_powers_of_two(int n)
{
	int i=1;
	long sum=0;
	int cnt=0;

	while (cnt < n) {
		printf("2 power %d = %d\n", cnt, i);
		sum += i;
		i = i << 1;
		cnt++;
	}

	printf("Sum of first n powers of two is: %ld\n", sum);
}

(2<<N)-2

( 2^(N+1) ) - 2

An implementation without Bit shift

public int  bruteForceIt(int x){
		
		int j = 1;
		int y = 0;
		for (int i = 1; i <= x; i++) {
			 if(x > 0){
				 
				j = j*2;
				y += j;
				System.out.println(j);
			}	
		}
		if(x == 0){
			System.out.println("The Sum of Powers is" +  1);
			return 1;
		}
			
		System.out.println("The Sum of Powers is" +  y);
		return j;
	}

Using bit shift to find the powers

public static void bitShitForSumower(int j){
		
		int x = 1;
		int y = 0;
		for (int i = 0; i < j; i++) {
			//bit shift 
			x = x<<1;
			y += x;
			System.out.println("x = " + x );
		}
		
		if( j == 0) System.out.println("The Some is 1"); 
		else { System.out.println("The some is " + y);}
			
	}

Hi guys, thanks for all the answers. I was able to produce brute force solution by calculating and adding powers inside the loop without bit manipulation but my solution didn't pass timing tests, correctness tests were ok.
Is it something obvious that if i is some power of two and you do i << 1, then power gets increased by 1? It looks very elegant but I have no idea how you come up with this solution.

This can run in O(log x). This can run for any power function.

public static int power(int y, int x){
if(x == 1)
	return x;
if(x==0)
	return 1;
int prod = 1;
if(x % 2 != 0){
	prod *= 2;
	x--;
}
return power(y*y, x/2);

}

public static int power(int x){
	return power(2,x);
}

How about this?

function sum_power2($power)
{
	$total= 0;$i=0;
	while($i<=$power)
	{
		$total = $total+ power_2($i);		
		$i++;		
	}
	echo $total;		
}
function power_2($i)
{
	//base case
	if ($i==0) return 1;
	if($i==1) return 2;
	return power_2($i-1)*2;
}

sum_power2(3);

How about this?

function sum_power2($power)
{
	$total= 0;$i=0;
	while($i<=$power)
	{
		$total = $total+ power_2($i);		
		$i++;		
	}
	echo $total;		
}
function power_2($i)
{
	//base case
	if ($i==0) return 1;
	if($i==1) return 2;
	return power_2($i-1)*2;
}

sum_power2(3);

Here is the formula : 2^(n+1) - 2

C# Solution :

private static int powerOfTwo(int power)
{
    int result = 1;

    for (int i = 0; i < power; i++)
        result = result * 2;

    return result;
}

private static int sumOfPowersOfTwo(int n)
{
    return powerOfTwo(n + 1) - 2;
}

If only I could understand the question..

int n = 10;
int sum = 0;
        for (int i = 0; i <= n; i++) {
            sum += power(i, 2);
        }

 private int power(int i, int p)  {
        if(p == 0) {
            return 1;
        }

        if (i == 0) {
            return 0;
        }
        else {
            return i * i;
        }
    }

Does this not work ?

unsigned long long sumP2(unsigned long long n) {

        return (1 << n ) - 1;
}