The sum of the digits in base ten of (10^{(4n2 +8)} +1)^{2}, where n is a positive integer, is:
- 4
- 4n
- 2 + 2n
- 4n^{2}
- n^{2} + n + 2
(10^{(4n2 +8)} +1) will look like this 1000…00001; that is, a number with a 1 on each end and 0’s in between.
If we square it, we have
10…01 | ||
x | 10…01 | |
10…01 | ||
+ | 10…010 … 0 | |
10…1..1…01 | or possibly | |
10…2…01 |
So the sum of the digits is 4, answer (a).
The 4n^{2} + 8 is a total red herring.