frac{x^2 - y^2}{1!} + frac{x^4 - y^4}{2!} + f | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) We know that the exponential series is given by $e^z = 1 + \frac{z}{1!} + \frac{z^2}{2!} + \frac{z^3}{3!} + \dots \infty$. Substituting $z = x^2$,

Vedclass · Accepted Answer

$e^{x^2} - e^{y^2}$

Vedclass · Answer

(B) We know that the exponential series is given by $e^z = 1 + \frac{z}{1!} + \frac{z^2}{2!} + \frac{z^3}{3!} + \dots \infty$. Substituting $z = x^2$,we get $e^{x^2} = 1 + \frac{x^2}{1!} + \frac{x^4}{2!} + \frac{x^6}{3!} + \dots \infty$. Similarly,substituting $z = y^2$,we get $e^{y^2} = 1 + \frac{y^2}{1!} + \frac{y^4}{2!} + \frac{y^6}{3!} + \dots \infty$. Subtracting the two series: $e^{x^2} - e^{y^2} = (1 - 1) + \frac{x^2 - y^2}{1!} + \frac{x^4 - y^4}{2!} + \frac{x^6 - y^6}{3!} + \dots \infty$. Thus,the given expression is equal to $e^{x^2} - e^{y^2}$.

$\frac{x^2 - y^2}{1!} + \frac{x^4 - y^4}{2!} + \frac{x^6 - y^6}{3!} + \dots \infty = $

Similar Questions

$1 + \frac{2^4}{2!} + \frac{3^4}{3!} + \frac{4^4}{4!} + \dots \infty = $

In the expansion of $\frac{a + bx + cx^2}{e^x}$,the coefficient of $x^n$ is

The sum of the series $\frac{1}{1 \times 2} + \frac{1 \times 3}{1 \times 2 \times 3 \times 4} + \frac{1 \times 3 \times 5}{1 \times 2 \times 3 \times 4 \times 5 \times 6} + \dots \infty$ is

In the expansion of $\frac{a + bx}{e^x}$,the coefficient of $x^r$ is

If $\cosh (x-\log 3)=\sinh x$,then $x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module