Câu hỏi của Nguyễn Thị Thanh Nhàn

Question

tính tổng:   S=3+3/2+3/22+.........+3/29

✓ ℍɠŞ_ŦƦùM $₦G ✓ · Accepted Answer

$S=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}$$=3\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)=3.\left(2-1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{2^2}+...+\frac{1}{2^8}-\frac{1}{2^9}\right)$$=3\left(2-\frac{1}{2^9}\right)=3.\frac{2^{10}-1}{2^9}=\frac{3\left(2^{10}-1\right)}{2^9}$ Câu trả lời: $S=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}$$=3\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)=3.\left(2-1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{2^2}+...+\frac{1}{2^8}-\frac{1}{2^9}\right)$$=3\left(2-\frac{1}{2^9}\right)=3.\frac{2^{10}-1}{2^9}=\frac{3\left(2^{10}-1\right)}{2^9}$

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