New prime in base 2

$2^2-1=11_2 \\ 2^3-1=111_2 \\ 2^4-1=1111_2$

As we can see, a pattern emerges. Now let's try to prove the pattern. There are many interesting ways to prove it. Here are 2 of them.

Algebraic way

We know that -

${ k }^{ n }-1=\left( k-1 \right) \left( { k }^{ n-1 }+{ k }^{ n-2 }+{ k }^{ n-3 }+\dots +{ k }^{ 2 }+k+1 \right)$

On plugging in 2, we get -

$\left( 2-1 \right) \left( { 2 }^{ 74,207,280 }+{ 2 }^{ 74,207,279 }+{ 2 }^{ 74,207,278 }+\dots +{ 2 }^{ 2 }+2+1 \right) \\ ={ 2 }^{ 74,207,280 }+{ 2 }^{ 74,207,279 }+{ 2 }^{ 74,207,278 }+\dots +{ 2 }^{ 2 }+2+1$

When we write this number in base 2, we get

$\underbrace { { 11111\dots 11 }_{ 2 } }_{ \text{74,207,281 ones}}$

Remember that the bases are 0 indexed. The sum is simply 74,207,281.

Number sense

$2^{74,207,281}$ is the smallest 74,207,282 digit number. Subtracting 1 from $2^{74,207,281}$ will form the largest 74,207,281 digit number, which is simply

$\underbrace { { 11111\dots 11 }_{ 2 } }_{ \text{74,207,281 ones} }$

Hence the sum is 74,207,281.