Question

how to solve this question?
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Answer 1

Answer:

n ( n + 1 ) / 2

Step-by-step explanation:

I see a couple of ways to go about this.  Choose whichever matches what you're doing in class and/or what you like.

Method 1  -  Pure algebra

$\displaystyle\frac{(x+x^2+x^3+\cdots+x^n)-n}{x-1}\\\\=\frac{(x-1)+(x^2-1)+(x^3-1)+\cdots+(x^n-1)}{x-1}\\\\=\frac{x-1}{x-1}+\frac{x^2-1}{x-1}+\frac{x^3-1}{x-1}+\cdots+\frac{x^n-1}{x-1}\\\\=1+(x+1)+(x^2+x+1)+\cdots+(x^{n-1}+\cdots+x+1)$

Now letting x tend to 1, this sum becomes

1 + 2 + 3 + ... + n  =  n ( n + 1 ) / 2

Method 2  -  L'Hospital's Rule

When the expression tends to the form 0/0, we can work instead with the form where the numerator and denominator are replaced by their derivatives.  If the result still tends to the form 0/0, we can do the same again.

$\displaystyle x+x^2+\cdots+x^n=x(1+x+\cdots+x^{n-1})=\frac{x(x^n-1)}{x-1}$

So...

$\displaystyle\lim_{x\rightarrow1}\frac{(x+x^2+x^3+\cdots+x^n)-n}{x-1}\\\\=\lim_{x\rightarrow1}\frac{x(x^n-1)-n(x-1)}{(x-1)^2}\\\\=\lim_{x\rightarrow1}\frac{x^{n+1}-x-nx+n}{(x-1)^2}\\\\=\lim_{x\rightarrow1}\frac{(n+1)x^n-1-n}{2(x-1)}\qquad\text{(by L'Hospital's Rule)}\\\\=\lim_{x\rightarrow1}\frac{n(n+1)x^{n-1}}{2}\qquad\text{(by L'Hospital's Rule again)}\\\\=\frac{n(n+1)}{2}$