First slide

Binomial theorem for positive integral Index

Question

If $a_{n} = 2^{2^{n}} + 1$ then for $n > 1$ , last digit of $a_{n}$ is

Moderate

A

5
B

7
C

3
D

4

Solution

For $n = 2, a_{n} = 2^{4} + 1 = 16 + 1 = 17$ . . Assume
That $a_{k} = 2^{2^{k}} + 1 = 10 m + 7$ where $k > 1$
and m is some positive integer.
Now, $\begin{array}{l} a_{k + 1} = 2^{2^{k + 1}} + 1 = {(2^{2^{k}})}^{2} + 1 \\ = (10 m + 6)^{2} + 1 = 10 (10 m^{2} + 12 m + 3) + 7 \end{array}$
Thus, last digit of $a_{n}$ is $7 \forall n > 1$

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