how many terms of the sequence 3^1/2 , 3, 3.3^1/2 ,.......... must be taken to make the sum 39+13.3^1/2?
Answers
The simplest way to figure this is out, is to group the “1, 2, 3”s in groups of, well, three. We know that 100 is not a multiple of 3, but 99 is. So we now know that the trio of “1, 2, 3”s will end at the 99th place, and the 100th place must be 1.
Great. So, we have 33 groups of “1, 2, 3”, plus a “1” leftover. There are 33 groups because there are 99 digits in the series divided up into groups of 3 each. 99 ÷ 3 = 33, so there are 33 groups.
Using algebra, we can visualize this series of numbers as “33 groups of ‘1 + 2 + 3’, plus 1”, or 33 ⨉ (1 + 2 + 3) + 1. This is easy to simplify:
33 ⨉ (1 + 2 + 3) + 1
= 33 ⨉ (6) + 1
= 198 + 1
= 199
So, the sum of digits “1, 2, 3, 1, 2, 3, 1, 2, 3,” repeating in such a fashion until there are 100 terms altogether, is 199.
This might not be the “correct” way to visualize and figure this out, but it’s one way, that’s for sure.
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