Find the sum of the products of $1,\frac{1}{2},\frac{1}{2^2},\dots,\frac{1}{2^{10}}$ taking two at a time.
My attempt is as follows:
$$S=1\cdot\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\dots\frac{1}{2^{10}}\right)+\frac{1}{2}\cdot\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\dots+\frac{1}{2^{10}}\right)\\+\frac{1}{2^2}\cdot\left(\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\dots\frac{1}{2^{10}}\right)+\dots+\frac{1}{2^9}\cdot\frac{1}{2^{10}}$$
$$T_n=\frac{1}{2^{n-1}}\cdot\left(\frac{1}{2^n}\cdot\left(\frac{1-\left(\frac{1}{2}\right)^{10-n+1}}{1-1/2}\right)\right)$$
$$S=\sum_{n=1}^{10}T_n=\sum_{n=1}^{10}\frac{1}{2^{n-1}}\cdot\left(\frac{1}{2^n}\cdot\left(\frac{1-\left(\frac{1}{2}\right)^{10-n+1}}{1-1/2}\right)\right)$$
$$S=\sum_{n=1}^{10}\frac{1}{2^{2n-2}}-\frac{1}{2^{2n-2}}\cdot\frac{1}{2^{10-n+1}}$$
$$S=\sum_{n=1}^{10}\left(1+\frac{1}{2^2}+\frac{1}{2^4}+\dots+\frac{1}{2^{18}}\right)-\sum_{n=1}^{10}\left(\frac{1}{2^{n+9}}\right)$$
$$S=\sum_{n=1}^{10}\left(1+\frac{1}{2^2}+\frac{1}{2^4}+\dots+\frac{1}{2^{18}}\right)-\sum_{n=1}^{10}\left(\frac{1}{2^{n+9}}\right)$$
$$S=\left(\frac{1-\left(\frac{1}{4}\right)^{10}}{1-\frac{1}{4}}\right)-\sum_{n=1}^{10}\left(\frac{1}{2^{n+9}}\right)$$
$$S=\left(\frac{4\cdot\left(4^{10}-1\right)}{4^{10}\cdot 3}\right)-\frac{1}{2^9}\cdot\left(\frac{1}{2}+\frac{1}{2^2}+\dots+\frac{1}{2^{10}}\right)$$
$$S=\left(\frac{4^{10}-1}{4^{9}\cdot 3}\right)-\frac{1}{2^9}\cdot\left(\frac{1}{2}\cdot\left(\frac{(1-\left(\frac{1}{2}\right)^{10}}{1-\frac{1}{2}}\right)\right)$$
$$S=\frac{4}{3}-\frac{1}{2^{18}\cdot 3}-\frac{1}{2^{19}}\cdot\left(2^{10}-1\right)$$
$$S=\frac{4}{3}-\frac{1}{2^{18}\cdot 3}-\frac{1}{2^9}+\frac{1}{2^{19}}$$
$$S=\frac{4}{3}+\frac{1}{2^{18}}\cdot\left(\frac{1}{2}-\frac{1}{3}\right)-\frac{1}{2^9}$$
$$S=\frac{4}{3}+\frac{1}{2^{18}}\cdot\frac{1}{6}-\frac{4}{2^{11}}$$
$$S=\frac{4}{3}+\frac{4}{2^{21}\cdot 3}-\frac{4}{2^{11}}$$
$$S=4\cdot\left(\frac{1}{3}+\frac{1}{2^{21}\cdot 3}-\frac{1}{2^{11}}\right)$$
But actual answer is $S=\dfrac{1}{3}+\dfrac{1}{2^{21}\cdot 3}-\dfrac{1}{2^{11}}$
I checked multiple times but not able to understand the mistake. What am I missing here.
Your answer is correct.
Note that $$(1+\frac {1}{2} +\frac {1}{4}...+\frac {1}{2^{10}})^2 =$$
$$(1+\frac {1}{4} +\frac {1}{4^2}...+\frac {1}{4^{10}}) + 2S$$
Where $S$ is the desired product.
Note that $$(1+\frac {1}{2} +\frac {1}{4}...+\frac {1}{2^{10}})^2 =$$
$$ (\frac{1-\frac {1}{2^{11}}} {1-\frac {1}{2}})^2 =$$
$$ 4+\frac {1}{2^{20}}-\frac {1}{2^8}$$ Also, $$(1+\frac {1}{4} +\frac {1}{4^2}...+\frac {1}{4^{10}})=$$
$$\frac {1}{3}(4-\frac {1}{2^{20}})$$
Thus $$S= 4(\frac {1}{3} + \frac {1}{3\times 2^{19}}-\frac {1}{2^{11}})$$
Which matches with your answer.