# Solving for K in integral calculus

$(6K+l)^{n+1}-l^{n+1}=60Kn+60K$

• Are you sure this is possible? I don't think it K can be computed explicitly.

• That's what i had thought, i was quite unsure though.

• I am pretty sure it is impossible.

• Math is so cool

To solve the equation $$(6K + l)^{n + 1} - l^{n + 1} = 60Kn + 60K$$ for K, we'll go through the steps:

1. Expand the terms on the left-hand side using the binomial theorem:
$$(6K + l)^{n + 1} = \binom{n+1}{0}(6K)^{n+1}l^0 + \binom{n+1}{1}(6K)^n l^1 + \binom{n+1}{2}(6K)^{n-1} l^2 + \ldots + \binom{n+1}{n}(6K)^0 l^{n+1}$$

2. Simplify the expression:
$$(6K + l)^{n + 1} = (6K)^{n+1} + (n+1)(6K)^n l + \binom{n+1}{2}(6K)^{n-1} l^2 + \ldots + \binom{n+1}{n} l^{n+1}$$

3. Now substitute this expansion into the original equation:
$$(6K)^{n+1} + (n+1)(6K)^n l + \binom{n+1}{2}(6K)^{n-1} l^2 + \ldots + \binom{n+1}{n} l^{n+1} - l^{n + 1} = 60Kn + 60K$$

4. Combine like terms and group all terms involving K on one side:
$$(6K)^{n+1} + (n+1)(6K)^n l + \binom{n+1}{2}(6K)^{n-1} l^2 + \ldots + \binom{n+1}{n} l^{n+1} - 60Kn - 60K = 0$$

This equation involves powers of K, and in most cases, it does not have a straightforward algebraic solution. To find an approximate solution for K, you can use numerical methods such as Newton's method or the bisection method. These methods involve making initial guesses for K and iteratively refining the solution until a desired level of accuracy is achieved.

• Uhhh, ok..

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