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amorica
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amorica
Can someone explain the quickest way to solve this?

2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 =??

sorry..but i forgot to write down the answers... but they were all in the form of 2^x


Since,

2^0 + 2^1 + 2^2 + ... + 2^n = 2^(n+1) -1

So,
1 + 2^0 + 2^1 + 2^2 + ... + 2^n = 2^(n+1) - 1 + 1

Ok then,
= 1 + 2^0 + 2^2 + ... + 2^8 = 2^(8+1) - 1 + 1
= 2^9



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