If you tutor math 12, you’ll have come across the premise<\/p>\n
1 + 1\/2 + 1\/4 + 1\/8 +….<\/p>\n
Before we go any further, a reminder on how to multiply fractions:<\/p>\n
(2\/3) x (4\/5) = 8\/15.\u00a0 You just multiply the two top numbers and put that number on top.\u00a0 Then you multiply the two bottom ones and put that number on the bottom.<\/p>\n
Now, back to 1 + 1\/2 + 1\/4 + 1\/8 + 1\/16+….(the terms continue forever)<\/p>\n
To generate the next term, you just multiply the previous one by, in this case, 1\/2.<\/p>\n
(1\/4) x (1\/2) = 1\/8<\/p>\n
1 + 1\/2 + 1\/4 + 1\/8 + 1\/16 + 1\/32 + 1\/64 +….<\/p>\n
In spite of the infinite\u00a0number of terms, the answer is 2.<\/p>\n
1 + 1\/2 + 1\/4 + 1\/8 + 1\/16 +1\/32 + 1\/64 + (the terms go on forever)….= 2.<\/p>\n
How can there be a sum of an unending number of terms?\u00a0 Well, there is a proof for it – but that doesn’t mean you have to believe it.\u00a0 Perhaps the best answer is that from a mathematical point of view, you can<\/em> add up an infinite number of terms.\u00a0 However, in this context,\u00a0two things have to be true:<\/p>\n 1) You must always multiply the current term by the same\u00a0factor to get the next one.<\/p>\n 2) That multiplying factor\u00a0must be less than 1.<\/p>\n So there isn’t a sum to 1+1\/2+1\/3+1\/4 + (the terms continue forever)….because the next number is not a constant multiple of the one before.<\/p>\n There isn’t a sum to 1 + 1 + 1 +\u00a0(the terms continue forever)….because we’re multiplying the current term by exactly 1 to get the next term.\u00a0 We must multiply by the same number each time, but it must also be less than one.<\/p>\n There is a sum to 1 + 2\/3 + 4\/9 + 8\/27 +….(infinite terms).\u00a0 Note that we are always multiplying by 2\/3 to get the next term.\u00a0 The sum is 3.<\/p>\n The formula for the sum is<\/p>\n S=a\/(1-r), where<\/p>\n a is the first term<\/p>\n r is the multiplying factor to get the next term (called the common ratio)<\/p>\n If you use a calculator, be sure to put the brackets around the denominator as shown in the formula.<\/p>\n Thanks for stopping by.<\/p>\n