Question

If S = (x+y)+(x² + xy + y?)+(x +x’y+xy? + y?)+...n terms,
w x? (x² - 1) y (x² - 1)
then prove that (x - y) Sn= x²(x^n -1) ÷ (x-1) - y² (y^n-1) ÷ (y-1)​

Answer 1

Answer:

$s = (x + y) + ({x}^{2} + + xy + {y}^{2} ) + {x}^{3} + ... \\ \\ \red{saprate \: terms}\\ = {y}^{0} (x + {x}^{2} + ... {x}^{n} ) + {y}^{1} (x + {x}^{2} + ... {x}^{n - 1} ) + .... \\ \\ \pink{ = \sum _{r = 0} ^{n} {y}^{r} (x + {x}^{2} + .... {x}^{ n - r} ) }\\ \\ = \sum {y}^{r} ( \frac{x( {x}^{n - r } - 1) }{x - 1} ) \\ \\ \green{break \: the \: summation \: } \\ \{note \: that \: terms \: not \: having \: r \: can \: be \: take \: out \: of \: sum \\ \\ = \frac{x}{x - 1} \sum {y}^{r} ( {x}^{n - r} - 1) \\ \\ = \frac{x}{x - 1} \bigg\{ \sum {y}^{r} . \frac{x {}^{n} }{ {x}^{r} } - \sum {y}^{r} \bigg\} \\ \\ = \frac{x}{x - 1} \bigg\{ {x}^{n} \sum(\frac{y {}^{} }{ {x}^{} } ) {}^{r} - \sum {y}^{r} \bigg\} \\ \\ = \frac{x}{x - 1} \bigg \{ {x}^{n} . \frac{(\frac{y {}^{} }{ {x}^{} } ) {}^{n} - 1 }{ \frac{y}{x} - 1} - \frac{ {y}^{n} - 1}{y - 1} \bigg \} \\ \\ = \frac{x}{x - 1} \{x. \frac{ {y}^{n} - {x}^{n} }{y - x} - \frac{ {y}^{n} - 1}{y - 1} \}$