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The factorial operation is encountered in many areas of mathematics notably in combinatorics.

1xn formula. Httpsyoutubems7il0vp6ha integral of 1x2. . We can write down the binomial expansion of 1xn as 1dfracn1x dfracnn 12x2 dfracnn 1n 23x3 this is true for all real.

The binomial theorem states that where n is a positive integer. Sum 31 1 2 4 8 16. Integral of 1x.

X 2 n 5 output. In elementary algebra the binomial theorem or binomial expansion describes the algebraic expansion of powers of a binomialaccording to the theorem it is possible to expand the polynomial x y n into a sum involving terms of the form ax b y c where the exponents b and c are nonnegative integers with b c n and the coefficient a of each term is a specific positive integer depending. Means n factorial and is equal to n n 1 2 1.

X 1 n 10 output. . If n is positive integer or positive rational number we can have the binomial expansion of math1x nmath as an infinite series as follows.

In general the rth number in the nth line is. The value of 0. Given an integer x the task is to print the series and find the sum of the series.

Integral of 1xn how to integrate it step by step. In mathematics the factorial of a positive integer n denoted by n is the product of all positive integers less than or equal to n. Nn 1.

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$1. \displaystyle{\int x^n dx = \dfrac{x^{n + 1}}{n + 1}} \\[3mm] \displaystyle{\int (ax + b)^n = \dfrac{1}{a} \dfrac{(ax + b)^{n + 1}}{n + 1}, n \neq -1} \\[3mm] 2.\displaystyle{\int \dfrac{1}{x} dx = \log x} \\[3mm] \displaystyle{\int \dfrac{1}{ax + b} dx = \dfrac{1}{a} \log (ax + b)} \\[3mm] 3. \displaystyle{\int e^x dx = e^x} \\[3mm] \displaystyle{\int e^{ax + b} dx = \dfrac{1}{a}e^{ax + b}} \\[3mm] 4. \displaystyle{\int a^x dx = \dfrac{a^x}{\log_e a}} \\[3mm] \displaystyle{\int a^{px + q} dx = \dfrac{1}{p} \dfrac{a^{px + q}}{\log_w a}}$

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$\\Coefficient \:of\:x^{n-2}\:is \:sum\:of\:roots\:taken\:2\:at\:a\:time=Required\:sum\\(1+2+3+4+.......+n)^2=1^2+2^2+3^2+4^2+.......n^2+2(Required\:sum)\\ \because sum\:of\:first\:n\:natural\:no\:\:is\:\:\frac{n(n+1)}{2}\\andsum\:of\:square\;of\;first\;n\:natural\:no\:is\:\frac{n(n+1)(2n+1)}{6}\\Solving\:above\:by\:using\:this\:formula\:you\:will\:get\:Coefficient \:of\:x^{n-2}$

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