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Calculus Examples
6∑k=0-k2-9
Step 1
Split the summation to make the starting value of k equal to 1.
6∑k=0-k2-9=6∑k=1-k2-9+0∑k=0-k2-9
Step 2
Step 2.1
Split the summation into smaller summations that fit the summation rules.
6∑k=1-k2-9=-16∑k=1k2+6∑k=1-9
Step 2.2
Evaluate 16∑k=1k2.
Step 2.2.1
The formula for the summation of a polynomial with degree 2 is:
n∑k=1k2=n(n+1)(2n+1)6
Step 2.2.2
Substitute the values into the formula and make sure to multiply by the front term.
(-1)(6(6+1)(2⋅6+1)6)
Step 2.2.3
Simplify.
Step 2.2.3.1
Simplify the numerator.
Step 2.2.3.1.1
Add 6 and 1.
-16⋅7(2⋅6+1)6
Step 2.2.3.1.2
Combine exponents.
Step 2.2.3.1.2.1
Multiply 6 by 7.
-142(2⋅6+1)6
Step 2.2.3.1.2.2
Multiply 2 by 6.
-142(12+1)6
-142(12+1)6
Step 2.2.3.1.3
Add 12 and 1.
-142⋅136
-142⋅136
Step 2.2.3.2
Simplify the expression.
Step 2.2.3.2.1
Multiply 42 by 13.
-1(5466)
Step 2.2.3.2.2
Divide 546 by 6.
-1⋅91
Step 2.2.3.2.3
Multiply -1 by 91.
-91
-91
-91
-91
Step 2.3
Evaluate 6∑k=1-9.
Step 2.3.1
The formula for the summation of a constant is:
n∑k=1c=cn
Step 2.3.2
Substitute the values into the formula.
(-9)(6)
Step 2.3.3
Multiply -9 by 6.
-54
-54
Step 2.4
Add the results of the summations.
-91-54
Step 2.5
Subtract 54 from -91.
-145
-145
Step 3
Step 3.1
Expand the series for each value of k.
-02-9
Step 3.2
Simplify.
Step 3.2.1
Raise 0 to the power of 2.
-0-9
Step 3.2.2
Multiply -1 by 0.
0-9
Step 3.2.3
Subtract 9 from 0.
-9
-9
-9
Step 4
Replace the summations with the values found.
-145-9
Step 5
Subtract 9 from -145.
-154