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## Yahoo! 2 years ago $\lim_{x \rightarrow 1}\frac{ 2x+3x^2........+(n+1)x^n - \frac{ n(n+3) }{ 2 } }{ x-1 }$

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1. Aryang

what is that written ? n(n+3)/2 ?

2. Aryang

please confirm ?

3. Yahoo!

Yes

4. shubhamsrg

did you try LH rule?

5. Aryang

seems legitimate to me..

6. shubhamsrg

there?

7. Yahoo!

Yup..I Tried...

8. Yahoo!

n is Constant Ryt ?

9. ash2326

Yes n is constant

10. AravindG

did u get the answer ?

11. Yahoo!

Nope

12. experimentX

|dw:1357836283297:dw|

13. experimentX

the derivative is $\frac{n x^{1+n}}{-1+x}+\frac{2 x \left(-1+x^n\right)}{-1+x}-\frac{x^2 \left(-1+x^n\right)}{(-1+x)^2}$

14. ash2326

We could find the sum of this first and then apply L'Hospital's $2x+3x^2+....(n+1)x^n$

15. experimentX

looks like we could use this formula directly http://planetmath.org/ArithmeticGeometricSeries.html

16. ash2326

$S=2x+3x^2+...(n+1)x^n$ $xS=2x^2+3x^3....(n+1)x^{n+1}$ $(1-x)S=2x+\underline{x^2+x^3+x^4....+x^n}-(n+1)x^{n+1}$ This is GP

17. experimentX

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

18. experimentX

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.

19. experimentX

this is equal to http://www.wolframalpha.com/input/?i=Sum [n%28n%2B1%29%2C+{n%2C+1%2C+n}]&dataset=&equal=Submit

20. Taufique

where is you confusion in this question??

21. Taufique

Apply L-hospital Rule...differentiate numerator and denominator with respect to variable which is x here ... |dw:1358023261199:dw| |dw:1358023336162:dw|

22. Yahoo!

2+ 6 + 12 ............... = n(n+1)

23. Taufique

my picture is not clear here.. now |dw:1358023498420:dw|

24. Taufique

|dw:1358023535161:dw|

25. Taufique

|dw:1358023576877:dw|

26. Taufique

|dw:1358023654085:dw|

27. Yahoo!

got it..:)

28. Taufique

solve it..and find L

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