Finding the nth Term of a Sequence

The problem asks us to find the formula for the n^th term of the given sequence: 25, -125, 625, -3125, …….

Let's analyze the pattern in the sequence:

• The first term is 25.
• The second term is -125.
• The third term is 625.
• The fourth term is -3125.

Analyzing the Numerical Values

Ignoring the signs for a moment, let's look at the absolute values of the terms:

• $|25| = 25 = 5^2$
• $|-125| = 125 = 5^3$
• $|625| = 625 = 5^4$
• $|-3125| = 3125 = 5^5$

We can see a clear pattern in the numerical values: the k^th term's absolute value is $5^{k+1}$. For the n^th term, the numerical part will be $5^{n+1}$.

Analyzing the Signs

Now let's look at the signs of the terms:

• Term 1 (n=1): Positive (+)
• Term 2 (n=2): Negative (-)
• Term 3 (n=3): Positive (+)
• Term 4 (n=4): Negative (-)

The signs alternate, starting with positive for n=1. An alternating sign can be represented using powers of -1.

• If the sign starts positive, we can use $(-1)^{n-1}$ or $(-1)^{n+1}$ (since n-1 and n+1 have the same parity difference).
• Let's check $(-1)^{n-1}$:
  • For n=1: $(-1)^{1-1} = (-1)^0 = 1$ (Positive)
  • For n=2: $(-1)^{2-1} = (-1)^1 = -1$ (Negative)
  • For n=3: $(-1)^{3-1} = (-1)^2 = 1$ (Positive)
  • For n=4: $(-1)^{4-1} = (-1)^3 = -1$ (Negative)
  This matches the required alternating sign pattern.

Combining Numerical and Sign Patterns

To get the n^th term, we combine the numerical part ($5^{n+1}$) and the sign part ($(-1)^{n-1}$).

So, the formula for the n^th term ($a_n$) is given by:

\(a_n = (-1)^{n-1} \times 5^{n+1}\)

Verification of the nth Term Formula

Let's test this formula for the first few terms:

• For n=1: \(a_1 = (-1)^{1-1} \times 5^{1+1} = (-1)^0 \times 5^2 = 1 \times 25 = 25\). This matches the first term.
• For n=2: \(a_2 = (-1)^{2-1} \times 5^{2+1} = (-1)^1 \times 5^3 = -1 \times 125 = -125\). This matches the second term.
• For n=3: \(a_3 = (-1)^{3-1} \times 5^{3+1} = (-1)^2 \times 5^4 = 1 \times 625 = 625\). This matches the third term.
• For n=4: \(a_4 = (-1)^{4-1} \times 5^{4+1} = (-1)^3 \times 5^5 = -1 \times 3125 = -3125\). This matches the fourth term.

The formula \(a_n = (-1)^{n-1} 5^{n+1}\) correctly generates the terms of the given sequence.

Comparing with Options

Let's compare our derived formula with the given options:

\(a_n = (-1)^{n-1} 5^{n+1}\)

• Option 1: \(a_n = (-5)^{2n - 1}\)
• Option 2: \(a_n = (-1)^{2n} 5^{n+1}\)
• Option 3: \(a_n = (-1)^{2n - 1} 5^{n+1}\)
• Option 4: \(a_n = (-1)^{n-1} 5^{n+1}\)

Option 4 is identical to our derived formula.

Conclusion

The n^th term of the sequence 25, -125, 625, -3125, ……. is given by the formula \(a_n = (-1)^{n-1} 5^{n+1}\).

Sequence and nth Term Formula Revision

Additional Information on Alternating Sequences

A sequence where the terms alternate between positive and negative signs is called an alternating sequence. The alternating part is typically represented by powers of -1.

• If the first term is positive and the signs alternate, the sign part is often $(-1)^{n-1}$ or $(-1)^{n+1}$.
• If the first term is negative and the signs alternate, the sign part is often $(-1)^n$ or $(-1)^{n-2}$.

The given sequence is also a type of geometric sequence where the common ratio alternates. The ratio between consecutive terms is $\frac{-125}{25} = -5$, $\frac{625}{-125} = -5$, $\frac{-3125}{625} = -5$. This is a geometric sequence with first term \(a = 25\) and common ratio \(r = -5\). The general formula for a geometric sequence is \(a_n = a \times r^{n-1}\).

Let's check if \(a_n = 25 \times (-5)^{n-1}\) matches our derived formula:

\(25 \times (-5)^{n-1} = 5^2 \times ((-1) \times 5)^{n-1} = 5^2 \times (-1)^{n-1} \times 5^{n-1}\)

\(= (-1)^{n-1} \times 5^2 \times 5^{n-1} = (-1)^{n-1} \times 5^{2 + (n-1)} = (-1)^{n-1} \times 5^{n+1}\)

Yes, the formula matches. This confirms our result from analyzing the pattern term by term.