## Loser66 one year ago What is wrong with this? $$1 + x+x^2 +\cdots +x^n = \sum_{j =0}^n x^j$$ Hence $$1+x +x^2 +\cdots +x^{n-1}\\= x^{n-1}+1+x +x^2+\cdots +x^n = x^{n-1}+\sum_{j=0}^n x^j = x^{n+1}+\dfrac{x^n -1}{x -1}$$ continue on comment Please, explain me.

1. Loser66

$$=\dfrac{(x-1)(x^{n+1})+(x^n-1)}{x-1}\\=\dfrac{x^{n+2}-x^{n+1}+x^n -1}{x-1}$$

2. Loser66

While when we go directly from $$1+x + x^2 +\cdots + x^{n-1}= \sum_{j =0}^n x^{j -1}=\dfrac{x^n -1}{x-1}$$

3. Loser66

@ganeshie8

4. ganeshie8

looks there are several typoes with - and + in the main question could you pls double check

5. Loser66

$1+x+x^2 +\cdots+x^n =\sum_{j=0}^n x^j =\dfrac{x^{n+1}-1}{x-1}$ $1+x+x^2 +\cdots +x^n+x^{n-1} = (\sum_{j=0}^n x^j) + x^{n-1}\\=\dfrac{x^{n+1}-1}{x-1}+x^{n-1}$ Simplify it, I get $$\dfrac{(x^{n+1}-1)+(x^{n-1}(x-1))}{x-1}$$ open parentheses of the numerator: $\dfrac{x^{n+1} -1+x^n -x^{n-1}}{x-1}$

6. Loser66

But if I apply directly the formula $\sum_{j=0}^n x^{j-1}= \dfrac{x^n -1}{x-1}$ it is different from the above. Why?

7. dan815

what do u mean

8. dan815

okay here is the geometric series formula

9. dan815

|dw:1441205497301:dw|

10. dan815

|dw:1441205576147:dw|

11. Loser66

yes

12. dan815

|dw:1441205592712:dw|

13. Loser66

I got that part, dan

14. dan815
15. Loser66

but that is the sum of 1 + x+ x^2 +....+x^n how about 1+x +x^2+.... +x^(n-1)?

16. dan815

1+x +x^2+.... +x^(n-1) = (X^n-1)/(x-1)

17. Loser66

hahaha.... I got you. I am dummy!! I interpret it as 1+x+x^2 +......+x^n then add one more term +x^(n-1) while it should be interpret as 1 + x+x^2 +....+x^(n-1) without the last term eeeeeeeeeeeh!! thank you very much, dan , the fake girl!!

18. dan815

yeah just remember if u are adding 1 more number to the 'sequence' 1+x+x^2 +......+x^n then + x^(n+1) not +x^(n-1) thats already in there

19. Loser66

yes, sir

20. dan815

no problem mam

21. Loser66

what do you mean by mam?

22. dan815

hahahhaha

23. Loser66

i do "hate" you man!!