Let f(x) = 1 + x + x^2 + x^3 + · · · + x^2011. Find f′(1)

Let f(x) = 1 + x + x^2 + x^3 + · · · + x^2011. Find f′(1)

would you say f(x)= 1+x^(2011-n)?

Note that, if f(x) = 1 + x + x^2 + x^3 + ... + x^2011, then:
f'(x) = 1 + 2x + 3x^2 + ... + 2011x^2010.

By letting x = 1:
f'(1) = 1 + 2(1) + 3(1)^2 + ... + 2011(1)^2010
= 1 + 2 + 3 + ... + 2011
= (2011)(1 + 2011)/2, since the sum of the first n positive integers is n(n + 1)/2
= 2,023,066.

I hope this helps!

You have a geometric series. The formula for that gives
f(x) = (1 - x^2012) / (1 - x)

The derivative can easily be computed from the quotient rule:
f'(x) = (-2012 x^2011)/(1 - x) + (1 - x^2012)/(1 - x)^2

However, the above is not actually valid for x=1, since my representation of f(x) has a divide by 0 error there. But that's alright--the derivative is valid everywhere else. Also, since f is a polynomial, it is continuously differentiable. In all, we can use my expression for f'(x) by computing the limit as x goes to 1. This may be done through some algebraic manipulation and application of L'Hopital's rule:

lim x->1 f'(x)
= lim x->1 (-2012 x^2011)/(1 - x) + (1 - x^2012)/(1 - x)^2 :: combine fractions into one term
= lim x->1 (2011 x^2012 - 2012 x^2011 + 1)/(x-1)^2 :: L'Hopital applies since this is 0/0
= lim x->1 (2011*2012 x^2011 - 2012*2011 x^2010) / [2(x-1)] :: L'Hopital applies again; 0/0
= lim x->1 (2011*2012*2011 x^2010 - 2012*2011*2010 x^2009) / 2 :: plug in 1
= [2011*2012*2011 (1) - 2012*2011*2010 (1)] / 2
= 2011*2012/2 * (2011-2010)
= 2011*2012 / 2
= 2023066

Incidentally, this is (2012 choose 2), which is just 1+2+...+2011, using the triangular number formula. However, my method generalizes to problems which don't happen to be susceptible to standard number theory formulas.

1 to any exponent is 1.

So, if it's 1^2011 (including 1^0, which is also 1), it's 1 + 1 + 1.... for 2012 ones.


2012 x 1 = 2012.


2012.

whats n? and no
f'(x)=1+2x+3x^2+...+2011x^2010
f'(1)=1+2+3+...+2011

oh my goodness i think im going to die of overthinking