Question

The value of $\sum_{k=1}^{10}(3k^2+2k-1)$ is

• A. 1120
• B. 1200
• C. 1230
• D. 1265
• E. 1255

Hint:

Formula Tip: Memorizing $\sum_{k=1}^{10}k = 55$ and $\sum_{k=1}^{10}k^2 = 385$ saves significant time during standardized tests!

Answer 1

Answer: (E)

Concept:
The summation operator is linear, meaning it can be distributed across addition and subtraction, and constants can be factored out. $$\sum (ak^2 + bk + c) = a\sum k^2 + b\sum k + \sum c$$

Step 1: Distribute the summation operator.
Split the given summation into three separate sums: $$S = 3\sum_{k=1}^{10}k^2 + 2\sum_{k=1}^{10}k - \sum_{k=1}^{10}1$$

Step 2: Recall the standard summation formulas.
Sum of squares: $\sum_{k=1}^{n}k^2 = \frac{n(n+1)(2n+1)}{6}$ Sum of integers: $\sum_{k=1}^{n}k = \frac{n(n+1)}{2}$ Sum of a constant: $\sum_{k=1}^{n}c = nc$

Step 3: Evaluate each sum for n = 10.
Evaluate the squares: $\sum_{k=1}^{10}k^2 = \frac{10(11)(21)}{6} = 5 \times 11 \times 7 = 385$ Evaluate the integers: $\sum_{k=1}^{10}k = \frac{10(11)}{2} = 5 \times 11 = 55$ Evaluate the constant: $\sum_{k=1}^{10}1 = 10 \times 1 = 10$

Step 4: Substitute the values back into the expression.
$$S = 3(385) + 2(55) - 10$$

Step 5: Calculate the final arithmetic result.
$$S = 1155 + 110 - 10$$ $$S = 1265 - 10$$ $$S = 1255$$ Hence the correct answer is (E) 1255.