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a(n)=-(1)/(16)(2)^(n-1) What is the 4^("th ") term in the sequence?

a(n)=−116(2)n−1 a(n)=-\frac{1}{16}(2)^{n-1} \newlineWhat is the 4th  4^{\text {th }} term in the sequence?

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Full solution

Q. a(n)=−116(2)n−1 a(n)=-\frac{1}{16}(2)^{n-1} \newlineWhat is the 4th  4^{\text {th }} term in the sequence?
  1. Step 11: Understand sequence formula: Understand the sequence formula and identify the term to find.\newlineThe sequence is defined by the formula a(n)=−(116)(2)n−1a(n) = -\left(\frac{1}{16}\right)(2)^{n-1}. We need to find the 4th4^{\text{th}} term, which means we will substitute nn with 44 in the formula.
  2. Step 22: Substitute n n with 4 4 : Substitute n n with 4 4 in the sequence formula.a(4)=−(116)(2)(4−1)a(4) = -\left(\frac{1}{16}\right)(2)^{(4-1)}
  3. Step 33: Calculate the exponent: Calculate the exponent part of the formula. 2(4−1)=23=82^{(4-1)} = 2^3 = 8
  4. Step 44: Multiply by −(116)-(\frac{1}{16}): Multiply the result of the exponentiation by −(116)-(\frac{1}{16}).\newlinea(4)=−(116)×8a(4) = -(\frac{1}{16}) \times 8
  5. Step 55: Perform the multiplication: Perform the multiplication to find the 4th4^{\text{th}} term.a(4)=−816a(4) = \frac{-8}{16}a(4)=−12a(4) = \frac{-1}{2}

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