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5 - 5/3 + 5/9 - 5/27 + ... = 5 + 5(-1/3)¹ + 5(-1/3)² + 5(-1/3)³ + ...

The required sum is an infinite GP with initial term a = 5, and common difference r = -1/3

As |r| < 1, the sum can be found from sum = a/(1 - r)

→ 5 - 5/3 + 5/9 - 5/27 + ...

= 5/(1 - (-1/3))

= 5/(1 + 1/3)

= 5/(4/3)

= 5 × 3/4

= 15/4

= 3¾

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7y ago

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