Converting Sigma Notation: Simplifying Binomial Theorem

[Saitama]

What do you get if you fill in x=0 in this formula?
f(x) = \frac{x(x+1)^{n+1}}{n+1}-\frac{(x+1)^{n+2}}{(n+1)(n+2)} - \frac {x^2} 2 + C

If i would fill x=0 in this formula i get
f(x) = \frac{0^{n+2}}{n+1}-\frac{0^{n+2}}{(n+1)(n+2)}+ C

Is it correct..?

 

[I like Serena]

Noooo. ;)
You need to fix the second term.

 

[Saitama]

Noooo. ;)
You need to fix the second term.

Ok fixed.
f(x) =-\frac{1^{n+2}}{(n+1)(n+2)}+ C

Is it ok..?

 

[I like Serena]

Yes...

 

[Saitama]

Yes...

But now what's the C?

 

[I like Serena]

But now what's the C?

Set the expression equal to zero (since we had f(0)=0) and solve C.

 

[Saitama]

Set the expression equal to zero (since we had f(0)=0) and solve C.

I set it to 0 and i get
C=\frac{1^{n+2}}{(n+1)(n+2)}

So therefore our final answer is:-
f(x)=\frac{x(x+1)^{n+1}}{n+1}-\frac{(x+1)^{n+2}}{(n+1)(n+2)} - \frac {x^2} 2 + \frac{1^{n+2}}{(n+1)(n+2)}

Substituting x=1, i get
f(1)=\frac{n.2^{n+2}-n^2-3n-4}{2(n+1)(n+2)}

Is it right..?
Can i simplify it further?

 

[I like Serena]

Is it right..?
Can i simplify it further?

You're expression for f(x) is right!

However your simplification after substitution of x=1 is wrong. :(

You can check by for instance filling in n=1.
(What should the result be for n=1 and for n=2? And does your expression match it?)

And actually I would recommend simplifying as little as possible.
Simplify the obvious things.
But do not try to simplify too much, since it opens up the door to mistakes.
And especially if the resulting expression is not really simpler, it's not worthwhile anyway.

 

[Saitama]

You're expression for f(x) is right!

However your simplification after substitution of x=1 is wrong. :(

You can check by for instance filling in n=1.
(What should the result be for n=1 and for n=2? And does your expression match it?)

And actually I would recommend simplifying as little as possible.
Simplify the obvious things.
But do not try to simplify too much, since it opens up the door to mistakes.
And especially if the resulting expression is not really simpler, it's not worthwhile anyway.

Thanks, i would correct it.
So now we are done with this question, Right..?

 

[I like Serena]

Thanks, i would correct it.
So now we are done with this question, Right..?

Hmm, you still didn't give the right answer...

But if you substitute x=1 without simplifying it, that would be a proper answer, so I guess we're done, with only a technicality left.