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Substitution

ntsolve:-n nt...

Question

ntsolve:-n nt(2+i)x square + (i-5)x - 2(i-1)=0n ntn

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t_{n} = \sum_{r = 0}^{n} \frac{1}{{({^{n} C}_{r})}^{k}}

S_{n} = \sum_{r = 0}^{n} \frac{r}{{({^{n} C}_{r})}^{k}}

k \in Z^{+}

{cos}^{- 1} (\frac{S_{n}}{n t_{n}})

\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i

x, y

.

\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i

(x, y) =

S_{n} = \frac{1}{n + 1} \sum_{i = 0}^{n} (\begin{matrix} n i \end{matrix})

2 S_{n + 1} - S_{n}

Q. Solve the equation

\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i; x, y \in R; i = \sqrt{- 1} .

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