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Question: $\lim_{x\to 1}\left[\frac{x+x^2+x^3+......+x^n-n}{x-1}\right]$...

lim⁡x→1[x+x2+x3+......+xn−nx−1]\lim_{x\to 1}\left[\frac{x+x^2+x^3+......+x^n-n}{x-1}\right]

Answer

n(n+1)2\frac{n(n+1)}{2}

Explanation

Solution

The given limit is lim⁡x→1[x+x2+x3+......+xn−nx−1]\lim_{x\to 1}\left[\frac{x+x^2+x^3+......+x^n-n}{x-1}\right].

Let f(x)=x+x2+x3+......+xnf(x) = x+x^2+x^3+......+x^n.
Evaluate f(x)f(x) at x=1x=1:
f(1)=1+12+13+......+1n=1+1+......+1f(1) = 1+1^2+1^3+......+1^n = 1+1+......+1 (n times) =n= n.

The given limit can be written as lim⁡x→1f(x)−nx−1\lim_{x\to 1}\frac{f(x)-n}{x-1}.
Since f(1)=nf(1)=n, the limit becomes lim⁡x→1f(x)−f(1)x−1\lim_{x\to 1}\frac{f(x)-f(1)}{x-1}.

This expression is the definition of the derivative of the function f(x)f(x) at x=1x=1, which is denoted as f′(1)f'(1).

To find f′(1)f'(1), we first find the derivative of f(x)f(x) with respect to xx.
f(x)=x+x2+x3+......+xnf(x) = x+x^2+x^3+......+x^n.
Using the power rule for differentiation, ddx(xk)=kxk−1\frac{d}{dx}(x^k) = kx^{k-1}, we differentiate each term:
f′(x)=ddx(x)+ddx(x2)+ddx(x3)+......+ddx(xn)f'(x) = \frac{d}{dx}(x) + \frac{d}{dx}(x^2) + \frac{d}{dx}(x^3) + ...... + \frac{d}{dx}(x^n)
f′(x)=1⋅x1−1+2⋅x2−1+3⋅x3−1+......+n⋅xn−1f'(x) = 1 \cdot x^{1-1} + 2 \cdot x^{2-1} + 3 \cdot x^{3-1} + ...... + n \cdot x^{n-1}
f′(x)=1⋅x0+2⋅x1+3⋅x2+......+n⋅xn−1f'(x) = 1 \cdot x^0 + 2 \cdot x^1 + 3 \cdot x^2 + ...... + n \cdot x^{n-1}
f′(x)=1+2x+3x2+......+nxn−1f'(x) = 1 + 2x + 3x^2 + ...... + nx^{n-1}.

Now, evaluate f′(x)f'(x) at x=1x=1:
f′(1)=1+2(1)+3(1)2+......+n(1)n−1f'(1) = 1 + 2(1) + 3(1)^2 + ...... + n(1)^{n-1}
f′(1)=1+2+3+......+nf'(1) = 1 + 2 + 3 + ...... + n.

This is the sum of the first nn positive integers. The formula for the sum of the first nn positive integers is n(n+1)2\frac{n(n+1)}{2}.
So, f′(1)=n(n+1)2f'(1) = \frac{n(n+1)}{2}.

Therefore, the value of the limit is n(n+1)2\frac{n(n+1)}{2}.