Question #275675
8∑(j=1)2^j





1
Expert's answer
2021-12-06T13:27:52-0500

Let us find the sum j=182j\sum\limits_{j=1}^82^j using the formula for the sum of geometric progression Sn=a1qn1q1.S_n=a_1\frac{q^n-1}{q-1}.

In our case, a1=2, q=2, n=8.a_1=2,\ q=2,\ n=8.

It follows that

j=182j=S8=228121=2255=510.\sum\limits_{j=1}^82^j=S_8=2\frac{2^8-1}{2-1}=2\cdot 255=510.


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