Prove that the sequence B (n) is a geometric progression if B (n) = 3/3 ^ 2-n. Find the sum of the first 6 terms.

| May 6, 2021 | education

Let us find the ratio of the n + 1-th term and B (n + 1) of the given sequence to the n-th term of B (n)

B (n + 1) / B (n) = (3 / 32- (n + 1)) / (3/32-n) = (3/32-n-1)) / (3/32-n) = (1/3) * (3/32-n) / (3/32-n) = 1/3.

Therefore, B (n + 1) = (1/3) * B (n).

This means that each term in this sequence, starting with the second, is equal to the previous term multiplied by 1/3.

Therefore, according to the definition of a geometric progression, this sequence is a geometric progression with the first term B (1) = 3 / 32-1 = 1 and the denominator q = 1/3.

Using the formula for the sum of the first n terms of the geometric progression Sn = B (1) * (1 – qⁿ) / (1 – q) for n = 6, we find the sum of the first 6 members of this sequence:

S6 = 1 * (1 – (1/3) 6) / (1 – 1/3) = (1 – 1/729) / (2/3) = (728/729) / (2/3) = ( 728/729) * 3/2 = 364/243.

Answer: The sum of the first 6 members of this sequence is 364/243.



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