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c(n)=-6(-(1)/(3))^(n-1)
What is the 
2^("nd ") term in the sequence?

c(n)=6(13)n1 c(n)=-6\left(-\frac{1}{3}\right)^{n-1} \newlineWhat is the 2nd  2^{\text {nd }} term in the sequence?

Full solution

Q. c(n)=6(13)n1 c(n)=-6\left(-\frac{1}{3}\right)^{n-1} \newlineWhat is the 2nd  2^{\text {nd }} term in the sequence?
  1. Identify the term: Identify the term of the sequence we need to find.\newlineWe are asked to find the second term in the sequence, which means we need to find c(2)c(2).
  2. Substitute the value: Substitute the value of nn into the sequence formula.\newlineThe formula for the sequence is c(n)=6(13)(n1)c(n)=-6(-\frac{1}{3})^{(n-1)}. To find the second term, we substitute n=2n=2 into the formula.\newlinec(2)=6(13)(21)c(2)=-6(-\frac{1}{3})^{(2-1)}
  3. Simplify the exponent: Simplify the exponent.\newlineCalculate the value of (13)(21)(-\frac{1}{3})^{(2-1)}, which is (13)1(-\frac{1}{3})^{1}.\newline(13)1=13(-\frac{1}{3})^{1} = -\frac{1}{3}
  4. Multiply the result: Multiply the result by 6-6.\newlineNow we multiply 13-\frac{1}{3} by 6-6 to get the second term.\newlinec(2)=6×(13)c(2) = -6 \times (-\frac{1}{3})
  5. Perform the multiplication: Perform the multiplication to find the second term. c(2)=6×(13)=63c(2) = -6 \times (-\frac{1}{3}) = \frac{6}{3}
  6. Simplify the fraction: Simplify the fraction. 63\frac{6}{3} simplifies to 22. c(2)=2c(2) = 2

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