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Complete the recursive formula of the arithmetic sequence

{:[-3","-1","1","3","dots],[b(1)=],[b(n)=b(n-1)+]:}

Complete the recursive formula of the arithmetic sequence\newline3,1,1,3,b(1)=b(n)=b(n1)+ \begin{array}{l} -3,-1,1,3, \ldots \\ b(1)=\square \\ b(n)=b(n-1)+\square\\ \end{array}

Full solution

Q. Complete the recursive formula of the arithmetic sequence\newline3,1,1,3,b(1)=b(n)=b(n1)+ \begin{array}{l} -3,-1,1,3, \ldots \\ b(1)=\square \\ b(n)=b(n-1)+\square\\ \end{array}
  1. Identify Pattern: Identify the pattern in the sequence to determine the common difference. The sequence given is 3,1,1,3,ext...-3, -1, 1, 3, ext{...}. To find the common difference, subtract the first term from the second term: 1(3)=1+3=2-1 - (-3) = -1 + 3 = 2. The common difference is 22.
  2. Recognize Formula: Recognize that a recursive formula for an arithmetic sequence has the form b(n)=b(n1)+db(n) = b(n-1) + d, where dd is the common difference. We have already determined that the common difference dd is 22.
  3. Write Recursive Formula: Write the recursive formula using the common difference and the first term of the sequence. The first term b(1)b(1) is given as 3-3. Therefore, the recursive formula is b(n)=b(n1)+2b(n) = b(n-1) + 2.
  4. Check Formula: Check the recursive formula by applying it to find the second term from the first term. Using b(n)=b(n1)+2b(n) = b(n-1) + 2, we get b(2)=b(1)+2=3+2=1b(2) = b(1) + 2 = -3 + 2 = -1, which matches the second term of the sequence.

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