Question

i need help with simplifying this problem

Answer 1

Ish, this is going to take a while. But it's simple. All you have to do is multiply is all out.

Answer 2

it simplifies down to n(n^2+6n+17)/3

Answer 3

i tried but i keep getting it wrong

Answer 4

Now you need to multiply (n^2+6n+17) by n, because you haven't finished FOILing.

Answer 5

its a sigma notation problem

Answer 6

yeah this is complicated

Answer 7

Okay, so I don't know what that is, I'm sorry. haha @KendrickLamar2014 wanna take it from here?

Answer 8

he left already

Answer 9

Yeah, I'm really sorry. I haven't even studied that yet, haha. @ganeshie8 you're smart, wanna help this guy out?

Answer 10

please @ganeshie8

Answer 11

i dont know what to do

Answer 12

whats the exact question ? if possible could u please take a screenshot and attach ?

Answer 13

i dont know how to do that

Answer 14

@ganeshie8

Answer 15

i am stuck on that

Answer 16

start by factoring out n

Answer 17

\[\large \dfrac{n(n+1)(2n+1)}{6} + \dfrac{3n(n+1)}{2} + 4n\]

Answer 18

\[\large n\left[ \dfrac{(n+1)(2n+1)}{6} + \dfrac{3(n+1)}{2} + 4\right]\]

Answer 19

get common denominator

Answer 20

\[\large n\left[ \dfrac{(n+1)(2n+1)}{6} + \dfrac{9(n+1)}{6} + \dfrac{24}{6}\right]\]

Answer 21

add up

Answer 22

\[\large n\left[ \dfrac{(n+1)(2n+1) + 9(n+1) + 24}{6} \right]\]

Answer 23

next expand the binomials in numberator

Answer 24

\[\large n\left[ \dfrac{2n^2+3n+1 + 9n+9 + 24}{6} \right]\]

Answer 25

\[\large n\left[ \dfrac{2n^2+12n+34}{6} \right]\]

Answer 26

finally factor out a 2 and cancel

Answer 27

oh man you did thank you!

Answer 28

i never thought about factoring out the n i am so stupid

Answer 29

np :) i never thought about factoring it too before seeing the answer u wanted