Find the remainder when the sum \(2^{2} + 2^{6} + 2^{10} + 2^{14} + \cdots + 2^{110}\) is divided by 10.
+118608 Since you have said this is a repeat you should have posted the address of the original post.
Can you find it and post it here?
Anyway, the answer will be the last digit which is the sum of the last digits.
4+4+ 4 etc
consider the series
2,6,10, .... 110
a=2 d=4 tn=110 find n
110=2+(n-1)*4
108/4=n-1
n=28
24*4=112
so the remainder will be 2
Note: I have not checked my answer.
