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\[\lim_{x \rightarrow 1}\frac{ 2x+3x^2........+(n+1)x^n - \frac{ n(n+3) }{ 2 } }{ x-1 }\]
what is that written ? n(n+3)/2 ?
did you try LH rule?
seems legitimate to me..
did u get the answer ?
|dw:1357836283297:dw|
the derivative is \[ \frac{n x^{1+n}}{-1+x}+\frac{2 x \left(-1+x^n\right)}{-1+x}-\frac{x^2 \left(-1+x^n\right)}{(-1+x)^2} \]
We could find the sum of this first and then apply L'Hospital's \[2x+3x^2+....(n+1)x^n\]
looks like we could use this formula directly http://planetmath.org/ArithmeticGeometricSeries.html
\[S=2x+3x^2+...(n+1)x^n\] \[xS=2x^2+3x^3....(n+1)x^{n+1}\] \[(1-x)S=2x+\underline{x^2+x^3+x^4....+x^n}-(n+1)x^{n+1}\] This is GP
the sum would be http://www.wolframalpha.com/input/?i=differentiate+%28x^2+%28x^n-1%29%2F%28x-1%29%29 that's you would get indeterminate form ... you can apply L'hopital rule after that
you should be getting \[ \frac{1}{3} n \left(2+3 n+n^2\right) \] or since you know you get indeterminate form ... you could apply L'hopital rule from the start. that's way easier.
this is equal to http://www.wolframalpha.com/input/?i=Sum [n%28n%2B1%29%2C+{n%2C+1%2C+n}]&dataset=&equal=Submit
where is you confusion in this question??
Apply L-hospital Rule...differentiate numerator and denominator with respect to variable which is x here ... |dw:1358023261199:dw| |dw:1358023336162:dw|
2+ 6 + 12 ............... = n(n+1)
my picture is not clear here.. now |dw:1358023498420:dw|