Calculating manually will take eons to solve it!

Call the sum $S$ . Subtract $S$ from $2S$ . Apply the formula for a finite geometric series. You're done!

A bit more work:
First, we have
$S=1 + 2(2) + 3(2^2) + 4(2^3) + ... +99(2)^{98} + 100(2)^{99}$ Therefore, $2S=2+2(2^2)+3(2^3)+4(2^4)+...+99(2^{99})+100(2^{100})$ So, evaluating $2S-S$ , we reach $2S-S=100(2^{100}) - (2^{99}+2^{98}+2^{97}+...+2^3+2^2+2+1)$ Now, we apply the formula for a finite geometric series: $2^{99}+2^{98}+2^{97}+...+2^3+2^2+2+1=\dfrac{1-2^{100}}{1-2}=2^{100}-1$ And finally, substituting our condensed geometric series, we find that $S=100(2^{100})-2^{100}+1=99(2^{100})+1$

Let $f(n)=1+2\cdot2+3\cdot2^2+\ldots+n\cdot2^{n-1}$ . We have that $f(n)=(n-1)\cdot2^n+1$ .

Proof: For $n=1$ , we have that our statement is true, since $f(1)=(1-1)\cdot2^1+1=0\cdot2^1+1=1$ .

So let's assume that $f(n)=(n-1)\cdot2^n+1$ . In this case:

$\begin{aligned}f(n+1)&=1+2\cdot2+3\cdot2^2+\ldots+(n+1)\cdot2^{n}\\ &=f(n)+(n+1)\cdot2^{n}\\ &=(n-1)\cdot2^n+1+(n+1)\cdot2^n\\ &=2n\cdot2^n+1\\ &=n\cdot2^{n+1}+1\end{aligned}$

So the statement holds true for all natural numbers.

Finally, $f(100)=99\cdot2^{100}+1$ .

First let x = 2.

Also, we let (1) S = 1 + 2x + 3x^2 + 4x^3 + ... + 100x^9. Then, (2) S * x = x + 2x^2 + 3x^3 + ... + 99x^99 + 100x^100.

Now, (1) - (2) => S(1 - x) = 1 + x + x^2 + x^3 + ... + x^99 - 100x^100. Notice the sequence: x + x^2 + x^3 + ... + x^99, using the formula on getting the sum of geometric sequence, we get [x(1 - x^99)]/(1 - x) and the substituting 2 to x, we now have S(1 - 2) = 1 + 2(1 - 2^99)/(1 - 2) - 100(2^99) => S(-1) = 1 - 2(1 - 2^99) - 100(2^100) => S(-1) = 1 + 2^100 - 2 - 100(2^100) => S(-1) = - 99(2^100) - 1 and so S = 99(2^100) + 1.