$a^{37} - a^{37}$ is exactly divisible by

a - b

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

a + b

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

a^{2} - b^{2}

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

All three

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

Open in App

Solution

The correct option is C All three
$∵ a^{37} - a^{37} = 0$
Since, 0 is divisible by every number except 0.
$∴$ Option D is correct.

Suggest Corrections

Similar questions

{(1 + x + 2 x^{2})}^{20} = 0 + a_{1} x + a_{2} x^{2} + . . . . . . + a_{40} x^{40}

a_{1} + a_{3} + a_{5} + . . . . . . + a_{37}

If $(1 + x + 2 x^{2})^{20} = a_{0} + a_{1} + a_{2} x^{2} + . . . + a_{40} x^{40}, t h e n a_{1} + a_{3} + a_{5} + . . . + a_{37} e q u a l s$ :

{(1 + x + x^{2})}^{20} = 40 \sum r = 0 a_{r} x^{r}

a_{1} + a_{3} + a_{5} + \dots a_{37}

$P (n) : a^{2 n} - b^{2 n} is divisible by a + b, \forall n ϵ N$

To prove P(n) using mathematical induction, the base case is

a, b, c

a^{2} = b^{2} + c^{2} - 2 b c \sqrt{1 - a^{2}}, b^{2} = c^{2} + a^{2} - 2 c a \sqrt{1 - b^{2}}, c^{2} = a^{2} + b^{2} - 2 a b \sqrt{1 - c^{2}}

a = c \sqrt{1 - b^{2}} + b \sqrt{1 - c^{2}}

Join BYJU'S Learning Program