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antiseptic1488 [7]
3 years ago
12

The sequence 2, 3, 5, 6, 7, 10, ... consists of all natural numbers which are neither perfect squares nor perfect cubes. Find th

e 75th term of this sequence.

Mathematics
1 answer:
BartSMP [9]3 years ago
7 0

Answer: The 75th term is 86

Step-by-step explanation:

Since the sequence consists of all natural numbers which are neither perfect squares nor perfect cubes, we can write out the next ninety numbers starting from 2, and then ruling out the squares and cubes. We then count to the 75th number to the desired term. This is shown in the attachment below.

We can as well sum the number of squares and cubes in the range with the number of the number of the desired term i.e

Number of squares = 8 ( 4,9,16,25,36,49,64,81)

Number of cubes = 2 (8 and 27; excluding 64 which is as well a square)

Desired term = 75

So, 8+2+75 = 85.

Therefore, the 85th term counting from 2 ( i.e 86) is the 75th term of the sequence.

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Then replacing x with √5 x (I'm assuming you mean √5 times x, and not √(5x)) gives

\displaystyle \cos\left(\sqrt 5\,x\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt5\,x\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^{2n}}{(2n)!}

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