1*2 +2*(2^2)+2*(2^3)............+100*(2^100)

Find the sum of series.

1+x+x^2+...+x^n = (x^(n+1)-1)/(x-1)

Now I think the beginning of your sequence should be

2*2^0 + 2*2^1 + 2*2^2 + ...

In that case, the sum is 2*(2^101-1)/(2-1)

On the other hand, there could be only one typo and the series is

1(2^1) + 2(2^2) + 3(2^3 + ... + 100(2^100)

In that case you have a hypergeometric series ...

I think I gotta go with Reiny. My answer is bogus because of the 100*2^100.

What was I thinking?

To find the sum of the series, we can first simplify the expression:

1 * 2 + 2 * (2^2) + 2 * (2^3) + ... + 100 * (2^100)

In general, each term in the series can be written as:

n * (2^n)

where n ranges from 1 to 100.

Next, let's factor out the common term of 2^n:

2^n * (1 + 2 + 2^2 + ... + 2^(n-1))

Notice that the term in the parentheses represents a geometric series with a common ratio of 2. The sum of a geometric series with a common ratio greater than 1 can be calculated using the formula:

Sum = a * (r^n - 1) / (r - 1)

where "a" is the first term, "r" is the common ratio, and "n" is the number of terms.

In our case, the first term (a) is 1, the common ratio (r) is 2, and the number of terms (n) is 100.

Using the formula, we can find the sum of each term in the parentheses:

Sum = 1 * (2^100 - 1) / (2 - 1)

Simplifying this expression gives us the sum of the geometric series within the parentheses. Finally, we multiply this sum by 2^n to account for each term in the original series:

Sum = 2^100 * (1 * (2^100 - 1) / (2 - 1))

Simplifying further:

Sum = 2^100 * (2^100 - 1)

Calculating this expression yields the sum of the series.