x-0-i-1-n-ix-i-1-

Question Number 195419 by CrispyXYZ last updated on 02/Aug/23

x \neq 0 . \underset{i = 1}{\sum^{n}} {i x}^{i - 1} = ?

Answered by MathedUp last updated on 02/Aug/23

x \neq 0 . \underset{i = 1}{\sum^{n}} {i x}^{i - 1} = ?

\underset{k = 1}{\sum^{m}} {k x}^{k - 1} = \frac{d}{d x} Σ x^{k} = \frac{d}{d x} \frac{x (x^{n} - 1)}{x - 1}

Commented by MathedUp last updated on 02/Aug/23

∴ \frac{m \cdot x^{m + 1} - (m + 1) \cdot x^{m} + 1}{{(x - 1)}^{2}}

Commented by MathedUp last updated on 02/Aug/23

Bonus if x \to 1 \underset{k = 1}{\sum^{m}} k \cdot x^{k - 1} = \underset{k = 1}{\sum^{m}} k

\lim_{x \to 1} \frac{m \cdot x^{m + 1} - (m + 1) x^{m} + 1}{{(x - 1)}^{2}} = \frac{m (m + 1)}{2}

and we can get Ramanujan Sum by

Change this Equation

So Funny : ⟩