Question

Limx→ -1 [1 + x + x² + ……….+ x 20 ]​

Answer 1

Answer:

Step-by-step explanation:

Please check the question

if the limit tends to 1 then,

$\lim_{x \to 1} [1+x+x^2+x^3+\ldots+x^{20} ]\\= \lim_{x \to 1} \frac{x^{21} -1}{x-1} (~\because~a+ar+ar^2+\ldots+ar^{n-1} = a\frac{r^n-1}{r-1}\text{ ,} \\\text{\quad \quad \quad \quad \quad \quad ~~~~~~where a is the first term of the gp \& r is the common difference)}\\= \lim_{x \to 1} \frac{x^{21} -1^{21}}{x-1} \\= 21\cdot (1)^{21-1} \quad (~\because~ \lim_{x \to a} \frac{x^{n} -a^{n}}{x-a} = n\cdot a^{n-1} )\\=21$

else,

$\lim_{x \to -1} [1+x+x^2+x^3+\ldots+x^{20} ]\\= \lim_{x \to -1} \frac{x^{21} -1}{x-1} \\= \frac{(-1)^{21}-1}{-1-1}\\= \frac{-1-1}{-1-1}\\=\frac{-2}{-2}\\=1$