Question

Simplify

Answer 1

\[n(n+1)(2n+1)/3-3n(n+1)+4n\]

Answer 2

Did you get this from something???

Answer 3

Or you want to write it in Sigma Notation???

Answer 4

Find \[\sum_{1}^{n}(2i^2-6i+4)\]

Answer 5

\[2\sum_{1}^{n}i^2-6\sum_{1}^{n}i+4=n(n+1)(2n+1)/3-3n(n+1)+4n \]

Answer 6

Take the LCD first and solve the parenthesis by distributive property..

Are you given with the answer???

Answer 7

Do you know the answer??

So, that I can know at what maximum I have to simplify this..

Answer 8

Answer
\[2n(n-1)(n-2)/3\]

Answer 9

Okay I try..

Answer 10

Firstly solve for last two terms:

\[\large -3n(n+1) + 4n = -3n^2 - 3n + 4n = n - 3n^2\]

Answer 11

Now take the LCD and then solve:

\[\frac{n(n+1)(2n+1)}{3} + n - 3n^2 = \frac{(n^2+n)(2n+1) + 3n - 9n^2}{3}\]

Answer 12

Now comparing with the answer you have given to me @strawberry_militia we have to solve for numerator..

Numerator should be equal to \(2n(n-1)(n-2)\)

Answer 13

So further solving for numerator:

\[(n^2+n)(2n+1) + 3n - 9n^2 = 2n^3 + 3n^2 + n + 3n - 9n^2 = 2n^3 - 6n^2+4n\]

Answer 14

Now you are just 2-3 steps away from your answer:

Take 2n common out of the \(\large 2n^3 - 6n^2 + 4n\)

\[2n(n^2 - 3n + 2)\]

Now can you do its factorization @strawberry_militia

Answer 15

After factorization you will get:

\[\large 2n(n-1)(n-2)\]

so the whole answer becomes:

\[\large \color{cyan}{\frac{2n(n-1)(n-2)}{3}}\]

Answer 16

cool, why did you choose to solve the last terms first?

Answer 17

This can ease the problem..

See if you have one term in fraction and other three terms in simple(not fraction)

Then firstly solve for the terms that are not in fraction:

Example:

\[\frac{3}{5} + 15 - 81 + 86\]

So, I think you will solve for other three terms..

Right??

Answer 18

86 - 81 + 15 = 5 + 15 = 20

\[\frac{3}{5} + 20 \implies \frac{103}{5}\]

Answer 19

\[n(n+1)(2n+1)/3+n-3n^2 = n(n+1)(2n+1)/3-(-n(n+1))???\]

Answer 20

why do you add n-3n^2 and not subtract??

Answer 21

See i have flipped them:

if you have:

\[\large -x + y\]

then you can write it as :

\[y - x\]

Answer 22

is it a must?

Answer 23

or your preference?

Answer 24

No..

Again it is to solve it is in easy way..

See,

Tell me:

What do you like the most??

adding or subtracting??

Answer 25

I think Addition Is easier than Subtraction..

We always do mistakes of - only..

Answer 26

cool.

Answer 27

Got??

Answer 28

yes, your legendary.

Answer 29

No, I am not..

I am noticing you for past few days, you are solving all the problems on Summation or Sigma Notation that is too of Sum Of Square Of n natural numbers, sum of first n natural number etc etc..

That is what I call Legendary..

So, you are Legendary...

Answer 30

its a struggle.

Answer 31

\[\large \color{green}{\textbf{Keep It Up..}}\]

Answer 32

\[\sum_{1}^{n} au _{n}=a \sum_{1}^{n}u _{n}\]

Answer 33

I do not see the point of of it.

Answer 34

What is \(u_n\)???

Answer 35

nth term of the series u

Answer 36

I know that I mean what is \(u\) ??

Answer 37

\[\sum_{u=1}^{n}\]

Answer 38

It is simply the sum of first n natural numbers;;

Answer 39

Here in place of i you are given u..

Can you do it??

Answer 40

I do can it, but I don't see the point of re writing it

Answer 41

You can write it as:

\[a \sum_{u = 1}^{n}(u) = a \times \frac{n(n+1)}{2}\]

Answer 42

my text made a big deal of showing now \[\sum_{1}^{n} au _{n}=a \sum_{1}^{n}u _{n}\]

Answer 43

I don't see the point. the first makes more sense to me.

Answer 44

Sorry I am not getting what you want to say..

Answer 45

don't worry my friend. Help me again soon.

Answer 46

Second is the correct one..

Because constants are to be taken out of the Sigma Notation..