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Evaluate (precalculus)

Evaluate (precalculus)-example-1
User Egidra
by
7.9k points

2 Answers

2 votes

Answer:


\boxed{\boxed{\sum_(i=1)^(11)(3i^2-8)=1430}}

Explanation:

The given expression is,


\sum_(i=1)^(11)(3i^2-8)


=\sum_(i=1)^(11)(3i^2)-\sum_(i=1)^(11)(8)


=3\sum_(i=1)^(11)(i^2)-\sum_(i=1)^(11)(8)

we know that,


\sum_(i=1)^(n)i^2=(n(n+1)(2n+1))/(6) and


\sum_(i=1)^(n)c=cn,\text{ Where c is a constant}

Hence,


3\sum_(i=1)^(11)(i^2)-\sum_(i=1)^(11)(8)=3\left((11(11+1)(2(11)+1))/(6)\right)-8(11)


=3\left((11(12)(23))/(6)\right)-8(11)


=3\left(506}\right)-8(11)


=1518-88


=1430

User Almaz
by
7.6k points
3 votes

Answer:

Option 4 which is 1,430.

Explanation:

Step 1

In this this step we will break up the terms in the summation symbol using the rules for manipulation summations.


\sum_(i=1)^(n)(aX^2_i+k)=\sum_(i=1)^(n)(aX)+\sum_(i=1)^(n)k\\\\\sum_(i=1)^(n)(aX^2_i+k)=a\sum_(i=1)^(n)X^2+nk.

This means that for our case, the sum can be simplified to be,


\sum_(i=1)^(11)(3i^2_i-8)=3\sum_(i=1)^(11)i^2+11*(-8)\\\\\sum_(i=1)^(11)(3i^2_i-8)=3\sum_(i=1)^(11)i^2-88

Step 2

In this step we use the formula for the sum of the first n squares to evaluate the first part of the sum in step 2. The formula for the sum of the first n squares is,


\sum_(i=1)^(n)n^2=(n(n+1)(2n+1))/(6) \\\\\implies\sum_(i=1)^(11)i^2=(11(11+1)(2* 11+1))/(6) =506

Step 3

In this step we use the result from step 1 to evaluate the sum.
3\sum_(i=1)^(11)i^2-88=3(506)-88=1,430.

The correct answer is Option 4,which is 1,430.



User Alieu
by
8.3k points

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