QUESTION IMAGE
Question
question 2 of 15, step 1 of 1 correct find the value of the sum \\(\sum_{j=1}^{28} (7j^2 - 3j + 5)\\). use a summation formula, if possible.
Step1: Split the summation
$\sum_{j=1}^{28} (7j^2 - 3j + 5) = 7\sum_{j=1}^{28} j^2 - 3\sum_{j=1}^{28} j + \sum_{j=1}^{28} 5$
Step2: Apply sum formulas
Recall:
$\sum_{j=1}^{n} j^2 = \frac{n(n+1)(2n+1)}{6}$, $\sum_{j=1}^{n} j = \frac{n(n+1)}{2}$, $\sum_{j=1}^{n} c = nc$
For $n=28$:
$7\cdot\frac{28(28+1)(2\cdot28+1)}{6} - 3\cdot\frac{28(28+1)}{2} + 5\cdot28$
Step3: Calculate each term
First term:
$7\cdot\frac{28\cdot29\cdot57}{6} = 7\cdot\frac{28\cdot29\cdot19}{2} = 7\cdot14\cdot29\cdot19 = 7\cdot14\cdot551 = 98\cdot551 = 53998$
Second term:
$3\cdot\frac{28\cdot29}{2} = 3\cdot14\cdot29 = 42\cdot29 = 1218$
Third term:
$5\cdot28 = 140$
Step4: Combine terms
$53998 - 1218 + 140$
Step5: Compute final value
$53998 - 1218 = 52780$; $52780 + 140 = 52920$
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52920