Certainly! Let's derive the expression for the [tex]\((n+1)\)[/tex]-th term given that the [tex]\(n\)[/tex]-th term of the sequence is represented by [tex]\(\frac{(n+3)(n+4)}{2}\)[/tex].
1. Identify the original expression: The [tex]\(n\)[/tex]-th term of the sequence is given by:
[tex]\[
T_n = \frac{(n+3)(n+4)}{2}
\][/tex]
2. Substitute [tex]\(n+1\)[/tex] instead of [tex]\(n\)[/tex]: To find the [tex]\((n+1)\)[/tex]-th term, we need to substitute [tex]\(n+1\)[/tex] for [tex]\(n\)[/tex] in the original expression. Let's perform the substitution:
[tex]\[
T_{n+1} = \frac{((n+1) + 3)((n+1) + 4)}{2}
\][/tex]
3. Simplify the substituted expression: Next, we simplify the terms within the parentheses.
- [tex]\((n+1) + 3\)[/tex] simplifies to [tex]\(n + 4\)[/tex].
- [tex]\((n+1) + 4\)[/tex] simplifies to [tex]\(n + 5\)[/tex].
Therefore, the expression transforms to:
[tex]\[
T_{n+1} = \frac{(n+4)(n+5)}{2}
\][/tex]
4. Result: The simplified expression for the [tex]\((n+1)\)[/tex]-th term in terms of [tex]\(n\)[/tex] is:
[tex]\[
T_{n+1} = \frac{(n+4)(n+5)}{2}
\][/tex]
So, the expression for the [tex]\((n+1)\)[/tex]-th term of the sequence, given that the [tex]\(n\)[/tex]-th term is [tex]\(\frac{(n+3)(n+4)}{2}\)[/tex], is [tex]\(\frac{(n+4)(n+5)}{2}\)[/tex].
