If an 22n 1 for n 1 what is the last digit of a451Solutio

If a_n = 2^2n + 1 for n > 1, what is the last digit of a_451?

Solution

2^ ( 2ⁿ+1) for n>1 and we are to find out the last digit of a₄₅₁;

it is important to note that 2¹=2 ; 2² = 4; 2³=8; 2⁴=16; 2⁵=32; 2⁶=64; Thus the last digit if a number 2^x where x is a natural number, depends upon the periodicity of x (which in this case is 4) thus if
x=4m then last digit = 6
x= 4m+1 then last digit =2
x=4m+2 then last digit = 4
x= 4m+3 then last digit = 8;
2⁴⁵¹ can be written as 2^[(2*225) +1] = 4²²⁵* 2¹;
Thus 2⁴⁵¹ when divided by 4 will give us a remainder = 2¹=2; Thus
2⁴⁵¹+ 1 can be written as 4k(2) +1;

a_n can be thus written as = 2^{(4k)2 +1} ; Thus the power of 2 is in the form of 4a+1 in this case its last disit will be =2;



Simply write 2⁴⁵¹ as 2²*2⁴⁴⁹ = 4(2⁴⁴⁹) Thus 2⁴⁵¹+1 can be written as 4(k) +1 where k=2⁴⁴⁹
Now 2^4k+1 will always have its last digit as =2;