$2^{3}$ + $2^{4}$ + $2^{5}$ + ....... + $2^{n}$ , where n > 0,
is always divisible by .

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A 2
$2^{3} = 8$
$2^{4} = 16$
$2^{5} = 32$
So, since 2 is an even number, $2^{3}$ , $2^{4}$ , $2^{5}$ ,...., $2^{n}$ (where n > 0) are even numbers.

We know that when two or more even numbers are added together, the sum is always an even number.

Therefore, $2^{3}$ + $2^{4}$ + $2^{5} + . . . + 2^{n}$ is always divisible by 2.

Suggest Corrections

Similar questions

2^{3}

2^{4}

2^{5}

2^{n}

2^{3}

2^{4}

2^{5}

2^{n}

n^{3} + 2 n^{2} - 5 n - 6,

+

n, n^{3} + 2 n

n

9^{2 n} - 4^{2 n}

Join BYJU'S Learning Program