Using the identity 1+x+x²+...+x^n = (x^(n+1)-1)/(x-1)

Using the identity $1+x+x^{2}+\cdots+x^{n} = \dfrac{x^{n+1}-1}{x-1}$ for $x \neq 1$, derive a formula for the sum $1+2x+3x^{2}+\cdots+nx^{n-1}$.

Step 1: State the identity

Both sides are polynomials in $x$ (after clearing the denominator) and agree for every $x \neq 1$: \[S(x) = 1 + x + x^{2} + \cdots + x^{n} = \frac{x^{n+1}-1}{x-1}, \qquad x \neq 1. \tag{1}\]

Two functions that coincide on an interval have the same derivative there, so we may differentiate both sides.

Step 2: Differentiate the left-hand side

Term by term, using $\bigl(x^{k}\bigr)' = kx^{k-1}$: \[S'(x) = \frac{d}{dx}\bigl(1 + x + x^{2} + \cdots + x^{n}\bigr) = 1 + 2x + 3x^{2} + \cdots + nx^{n-1}. \tag{2}\]

This is precisely the sum we are asked to evaluate.

Step 3: Differentiate the right-hand side

Apply the quotient rule with $u = x^{n+1}-1$, $u' = (n+1)x^{n}$, $v = x-1$, $v' = 1$: \[\frac{d}{dx}\!\left(\frac{x^{n+1}-1}{x-1}\right) = \frac{(n+1)x^{n}(x-1) - \bigl(x^{n+1}-1\bigr)}{(x-1)^{2}}.\]

Expanding the numerator: $(n+1)x^{n+1} - (n+1)x^{n} - x^{n+1} + 1 = nx^{n+1} - (n+1)x^{n} + 1$.

so that

\[ \frac{d}{dx}\!\left(\frac{x^{n+1}-1}{x-1}\right) = \frac{nx^{n+1}-(n+1)x^{n}+1}{(x-1)^{2}}. \tag{3} \]

Step 4: Equate (2) and (3)

Since the two sides of (1) are equal for all $x \neq 1$, their derivatives are equal as well.

Step 5: Verification

\[n = 2,\ x = 2:\quad \text{left} = 1 + 2\cdot 2 = 5; \quad \text{right} = \frac{2\cdot 8 - 3\cdot 4 + 1}{1} = \frac{16-12+1}{1} = 5 \;\;\checkmark\]

\[n = 3,\ x = -1:\quad \text{left} = 1 - 2 + 3 = 2; \quad \text{right} = \frac{3\cdot 1 - 4\cdot(-1) + 1}{4} = \frac{3+4+1}{4} = 2 \;\;\checkmark\]