Math, asked by snowwhiteaug18, 1 month ago

A geometric series has a common ratio of (-2) and the first term is 3.
Show that the sum of the first eight positive terms of the series is 65 535.

Answers

Answered by user0888
11

\huge\boxed{\textsf{Note}}

Sequence: Numbers organized in an order.

Series: Sum of the terms of a sequence.

\huge\boxed{\textsf{Solution}}

A geometric sequence (here, \{a_{n}\}) has a common ratio of -2 and the first term is 3.

\implies a_{n}=3\cdot (-2)^{n-1}

If the exponent is even, that term will be positive.

Put 2n-1 in place of n,

\implies a_{2n-1}=3\cdot (-2)^{2n-2}

\implies a_{2n-1}=3\cdot 4^{n-1}

Let us define the new sequence that only consists of positive terms. (Here, \{b_{n}\}) The sequence is defined as,

\implies b_{n}=3\cdot 4^{n-1}

\huge\boxed{\textsf{Note}}

Finite Geometric Series

\boxed{\displaystyle{\sum^{n}_{k=1} a_{k}}=\dfrac{a(r^{n}-1)}{r-1} }

The expression reads as "the sum of a_{k} as k goes from 1 to n."

Where,

  • a is the first term.
  • r is the common ratio. (r\neq 1)
  • n is the term number.

\huge\boxed{\textsf{Solution}}

We are given to find the first 8 terms.

The sequence \{b_{n}\} has

  • the first term of 3.
  • the common ratio of 4.

\hookrightarrow \ \displaystyle{\sum^{8}_{k=1}b_{k}}

\hookrightarrow \ =\dfrac{3(4^{8}-1)}{4-1}

\hookrightarrow\ =\dfrac{3(4^{8}-1)}{3}

\hookrightarrow \ =4^{8}-1

\hookrightarrow\ =65535

Hence shown.

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