8∑(j=1)2^j
Let us find the sum ∑j=182j\sum\limits_{j=1}^82^jj=1∑82j using the formula for the sum of geometric progression Sn=a1qn−1q−1.S_n=a_1\frac{q^n-1}{q-1}.Sn=a1q−1qn−1.
In our case, a1=2, q=2, n=8.a_1=2,\ q=2,\ n=8.a1=2, q=2, n=8.
It follows that
∑j=182j=S8=228−12−1=2⋅255=510.\sum\limits_{j=1}^82^j=S_8=2\frac{2^8-1}{2-1}=2\cdot 255=510.j=1∑82j=S8=22−128−1=2⋅255=510.
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