Infinite Roots
What is 2 0 + 2 0 + 2 0 + . . . . ? Write your answer as as a + b , where b a is equal to 2 0 + 2 0 + 2 0 + . . . . , and a and b are relatively prime
The answer is 6.
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5 solutions
Mine is a similar solution
That moment when you want to make your solution look different so you use 3 \boxed{} xD
Let's take,
2 0 + 2 0 + 2 0 + . . . . = x
We can also write it as,
2 0 + x = x 2 0 + x = x 2 x 2 − x − 2 0 = 0 x 2 − 5 x + 4 x − 2 0 = 0 x ( x − 5 ) + 4 ( x − 5 ) = 0 ( x − 5 ) ( x + 4 ) = 0
Therefore, the roots of the equation are, x = 5 and x = − 4 . As the second result is negative we can ignore it, So, the solution is 5 , or 1 5
And hence, a + b = 5 + 1 = 6
This looks exactly like @Sravanth Chebrolu 's solution above.
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Yeah, looks like he copied it.(see the dates)
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Yeah i saw that before I posted my earlier comment.
YES!! Jonathan, you can finally use L A T E X ! Solution: b a is equal to 1 5 , so a + b is equal to 5 + 1 or 6
Please explain how you got your answer.
As Sravanth Chebrolu has explained, the value is 5 . But do not forget that the question is asking for fraction form so the answer is NOT 5 . We assume the denominator to be 1 since we have a whole number so a + b = 5 + 1 = 6 .
Let's take, 2 0 + 2 0 + 2 0 + . . . . = x
We can also write it as, 2 0 + x = x 2 0 + x = x 2 x 2 − x − 2 0 = 0 x 2 − 5 x + 4 x − 2 0 = 0 x ( x − 5 ) + 4 ( x − 5 ) = 0 ( x − 5 ) ( x + 4 ) = 0
Therefore, the roots of the equation are, x = 5 and x = − 4 . As the second result is negative we can ignore it, So, the solution is 5 , or 1 5
And hence, a + b = 5 + 1 = 6