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Find the remainder when the sum \(2^{2} + 2^{6} + 2^{10} + 2^{14} + \cdots + 2^{110}\) is divided by 10.

 Oct 1, 2021
 #1
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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.

 Oct 2, 2021

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