All Math and no play makes Pete smart

Indeed, they can.

$2^{k+1}+ 2^{k+2}+.......+2^{k+10}= 1+2+3+.....+n$

$2^{k+2}(2^{10}-1)=n(n+1)$

We see that it works for $n=2^{10}-1, k=8$

Thus, the common sum = $\dfrac {2^{10} \times (2^{10} -1)}{2}= \boxed {523776}$

Great problem, thanks for sharing! Anyway, here's how I did it:

We have:

${ 2 }^{ x }+{ 2 }^{ x+1 }+\dots +{ 2 }^{ x+9 }=\frac { y(y+1) }{ 2 }$

We notice that the LHS is a geometric sequence. Simplifying: ${ 2 }^{ x+1 }({ 2 }^{ 10 }-1)=y(y+1)\\ 1023*{ 2 }^{ x+1 }=y(y+1)$

Notice that both the LHS and RHS consists of an odd number multiplied by an even number.

Thus we know that either $y$ or $y+1$ has ${ 2 }^{ x+1 }$ as a divisor. So we can re-write either $y$ or $y+1$ as ${ a*2 }^{ x+1 }$ Thus we have:

$1023*{ 2 }^{ x+1 }=a*{ 2 }^{ x+1 }({ a*2 }^{ x+1 }\pm 1)$

$\frac { 1023 }{ a } =({ a*2 }^{ x+1 }\pm 1)$

Both sides of the equation are integers, so $a$ is a divisor of $1023$ . So $a$ can be equal to: $1, 3, 11, 31, 33, 93, 341, 1023$

Re-arranging:

$\frac { 1023\pm a }{ { a }^{ 2 } } ={ 2 }^{ x+1 }$

We now try all possible cases ( $16$ cases, $8$ possible numbers for $a$ and $2$ possible variations of the $\pm$ sign) of $a$ and we find that $a$ can only equal $1$ for $x$ to be an integer, with the $\pm$ showing $+$ .

From the above equation, we then find that $x=9$ .

The answer, as required:

$1023*{ 2 }^{ x }=\boxed { 523776 }$