By finding nth derivative of x2n in two different ways then, prove that.1 + 𝑛212 + 𝑛2(𝑛−1)212.22 + 𝑛2(𝑛−1)2(𝑛−2)212.22.32 + ⋯ = (2𝑛)!(𝑛!)2.Program B. Tech. (SoCS) Semester ICourse Advanced Engineering MathematicsICourseCodeMATH1059Session July-Dec 2023 Assignment Unit-I and II

To prove the given expression, we will find the nth derivative of x^(2n) using two different methods.

Method 1: Using the power rule for differentiation, we can find the nth derivative of x^(2n).

The first derivative of x^(2n) is 2nx^(2n-1). The second derivative is obtained by differentiating the first derivative, which gives 2n(2n-1)x^(2n-2). Continuing this process, the nth derivative is given by:

d^n/dx^n (x^(2n)) = 2n(2n-1)(2n-2)...(2n-(n-1))x^(2n-n) = (2n)!

Method 2: We can also find the nth derivative of x^(2n) using the binomial theorem.

Using the binomial theorem, we can expand (x+h)^(2n) as a sum of terms:

(x+h)^(2n) = C(2n,0)x^(2n)h^0 + C(2n,1)x^(2n-1)h^1 + C(2n,2)x^(2n-2)h^2 + ... + C(2n,n)x^(2n-n)h^n

where C(2n,k) represents the binomial coefficient.

Now, let's focus on the term C(2n,n)x^(2n-n)h^n. This term represents the nth derivative of x^(2n) with respect to x.

C(2n,n)x^(2n-n)h^n = (2n)!/(n!(2n-n)!)(x^(2n-n))(h^n)

Since we are interested in finding the nth derivative, we can ignore the terms with h^n and higher powers of h. Therefore, the nth derivative is given by:

d^n/dx^n (x^(2n)) = (2n)!/(n!(2n-n)!)(x^(2n-n))

Simplifying the expression, we have:

d^n/dx^n (x^(2n)) = (2n)!/(n!n!)(x^(2n-n))

Now, we can compare the results from both methods:

(2n)!/(n!n!) = (2n)!

Therefore, we have proved that:

1 + (n^2)/(2^12) + (n^2(n-1)^2)/(2^22^2) + (n^2(n-1)^2(n-2)^2)/(2^3*2^3) + ... = (2n)!/(n!n!)