23-24-25- -2 n- where n -0is always divisible byA- 2B- 9C- 10D- 5

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

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Standard XII

Mathematics

Proof by mathematical induction

23+24+25+…… ....

Question

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

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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.

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23 + 24 + 25 + ....... + 2n, where n > 0, is always divisible by ____ .

The correct option is A 2 23=8 24=16 25=32 So, since 2 is an even number, 23, 24, 25,....,2n (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, 23 + 24 + 25+...+2n is always divisible by 2.

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