Divide a square with an area of 1 into two rectangles with an area of half Divide a square with an area of 1 into two rectangles with an area of 1 / 2, then divide the rectangle with an area of 1 / 2 into two rectangles with an area of 1 / 4, and then divide the rectangle with an area of 1 / 4 into two rectangles with an area of 1 / 8. In this way, try to use the law revealed by the graph to calculate: 1 / 2 + 1 / 4 + 1 / 8 + 1 / 6 + 1 / 32 + 1 / 128 =?

Divide a square with an area of 1 into two rectangles with an area of half Divide a square with an area of 1 into two rectangles with an area of 1 / 2, then divide the rectangle with an area of 1 / 2 into two rectangles with an area of 1 / 4, and then divide the rectangle with an area of 1 / 4 into two rectangles with an area of 1 / 8. In this way, try to use the law revealed by the graph to calculate: 1 / 2 + 1 / 4 + 1 / 8 + 1 / 6 + 1 / 32 + 1 / 128 =?

1/2+1/4+1/8+1/16+1/32+1/128
=1/2+(1/2)²+(1/2)³+(1/2)^4+(1/2)^5+(1/2)^6+(1/2)^7
=(1/2)×[1-(1/2)^7]÷(1-1/2)
=127/128
1 / 2 + 1 / 4 + 1 / 8 + 1 / 16 + 1 / 32 + 1 / 128 + ············· + (1 / 2) ^ n
=(1/2)×[1-(1/2)^n]÷(1-1/2)
=1-1/(2^n)
The n-th power infinity of N infinity 2 takes the reciprocal 1 / (2 ^ n) and approaches to 0
lim[1-1/(2^n)]=1
In the graph, no matter how many times it is divided, the sum of all rectangular areas is still equal to square area = 1