1.1.3 Evaluate $$ \sum_{P=1}^{n} T_{P} $$

Question

UpStudy AI · Accepted Answer

Let
$$
T_P = \frac{P(P+1)}{2}.
$$

Then, the sum is
$$
\sum_{P=1}^{n} T_P = \sum_{P=1}^{n} \frac{P(P+1)}{2} = \frac{1}{2} \sum_{P=1}^{n} \bigl( P^2 + P \bigr).
$$

Separate the summation:
$$
\frac{1}{2} \left( \sum_{P=1}^{n} P^2 + \sum_{P=1}^{n} P \right).
$$

Recall the formulas:
$$
\sum_{P=1}^{n} P^2 = \frac{n(n+1)(2n+1)}{6}
$$
and
$$
\sum_{P=1}^{n} P = \frac{n(n+1)}{2}.
$$

Substitute these into the expression:
$$
\frac{1}{2} \left( \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \right).
$$

Factor out $$ n(n+1) $$:
$$
\frac{1}{2} \left( n(n+1) \left[ \frac{2n+1}{6} + \frac{1}{2} \right] \right).
$$

Find a common denominator inside the brackets:
$$
\frac{2n+1}{6} + \frac{3}{6} = \frac{2n+4}{6} = \frac{2(n+2)}{6} = \frac{n+2}{3}.
$$

Thus, the sum becomes:
$$
\frac{1}{2} \cdot n(n+1) \cdot \frac{n+2}{3} = \frac{n(n+1)(n+2)}{6}.
$$

So, the evaluated sum is:
$$
\boxed{\frac{n(n+1)(n+2)}{6}}.
$$
The sum $$ \sum_{P=1}^{n} T_{P} $$ evaluates to $$ \frac{n(n+1)(n+2)}{6} $$.

1.1.3 Evaluate \( \sum_{P=1}^{n} T_{P} \)