pROGRESSIONS

I'll give hints only ---
1) a) By the time we reach the nth bracket, 1+2+3+...+(n-1) terms would have been used.

First term of nth bracket= n(n-1)2+1

b) Apply the previous logic here too.

2) arrange it like this =\begin{matrix} 1 & 0 &0 \\ 1& 0 & 1\\ 1& 0 & 2\\ . & . & .\\ 9&9 & 9 \end{matrix}

Now count how many times each digit appears in hundredth, tens and units places. Its not too difficult and the result will be symmetric for all digits 1-9. You dont need to rack your brains for 0 because it doesn't add to the sum.

I think the answer is 12600

3)
\begin{align*} S &=1+ \frac{1}{2}-\frac{1}{4}-\frac{1}{8}+\frac{1}{16}...\\ &=\left(1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}+... \right) + \left(\frac{1}{2}-\frac{1}{8}+\frac{1}{32}-\frac{1}{128}+... \right) \end{align*}

s=1+2*2+3*2²......100*2⁹⁹
2s= 2+2*2²+3*2³....100*2¹⁰⁰
s-2s=1+2+2²+2³.......-100*2¹⁰⁰
-s=1+2¹⁰⁰-12-1-100*2¹⁰⁰
s=99*2¹⁰⁰

what is s - sx ?

Find S=1+2x2+3x2²+4x2³.............100x2^99.

S means to find the sum.

x means to multiply.

Nice method Khilen! :)