Find the sum using the formulas for the sums of powers of integers ( problem attached )

Mathematics

1 answer:

Natasha2012 [34]3 years ago

7 0

Correct Answer: - 550

We can use the distributive property to distribute the summation to the variables and then get the constant out to apply the summation formulas as shown below in the attached image.

The equation editor does not have a summation symbol, so I have attached the solution using the image.

The correct answer to this question is -550. First option is the correct one.

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Factor this expression completely, then place the factors in the proper location on the grid. a^3y + 1

ivolga24 [154]

Answer:

$a^{3y} + 1 = (a^{y}+1 )^{3} - 3a^y(a^{y}+1)\\\\$

Step-by-step explanation:

We are to factorize the expression $a^{3y} + 1$ completely. To do this, we will apply the expression below;

The expression can be rewritten as $a^{3y} + 1^{3}$

To factorize the expression, we need to first factorize $(a^{y}+1 )^{3}$ first

$(a^{y}+1 )^{3} =(a^{y}+1 )(a^{y}+1 )^{2}\\= (a^{y}+1 )((a^y)^{2} } + 2a^{y} +1)\\= (a^y)^{3} +2(a^y)^{2}+a^y+( a^y)^{2}+2a^y+1\\(a^{y}+1 )^{3} = ((a^y)^{3} + 1) +2(a^y)^{2}+a^y+( a^y)^{2}+2a^y\\(a^{y}+1 )^{3} = ((a^y)^{3} + 1) +3(a^y)^{2}+3a^y\\$