write n with limit from 0 to infinity with the summation n on top and k=1 on the bottom (5+k(2/n))^10 (2/n) as a definite integral.

You want the integral to be something of the form

∫[a,b] f(x) dx

Now that (2/n) at the end indicates that we are dividing 2 into n parts, so we will have

∫[0,2] f(x) dx

Now we see that k*2/n inside the sum, so that is the value of x at each of the interval divisions, so I expect it to be

∫[0,2] (5+2x)^10 dx

Thank you!

To express the summation as a definite integral, we need to consider the limit as n approaches infinity. Let's break down the steps to convert the summation into a definite integral.

Step 1: Rewrite the expression inside the summation
Start by expanding the expression (5 + k(2/n))^10 using the binomial theorem.

(5 + k(2/n))^10 = (10C0)(5^10) + (10C1)(5^9)(k(2/n)) + (10C2)(5^8)(k(2/n))^2 + ...

Step 2: Replace the summation by the integral
Now, let's express the sum as an integral by dividing each term by n and multiplying by 2/n. This will allow us to convert the summation into an integral.

= Σ([(10C0)(5^10)/n^10] + [(10C1)(5^9)(k(2/n))/n^9] + [(10C2)(5^8)(k(2/n))^2/n^8] + ...)
= Σ([(10C0)(5^10)/n^10] * (2/n)^0) + [(10C1)(5^9)(k(2/n))/n^9] * (2/n)^1 + [(10C2)(5^8)(k(2/n))^2/n^8] * (2/n)^2 + ...

Step 3: Simplify the expression
Since n is approaching infinity, the terms containing n in the denominator will tend towards zero. Additionally, we can express the binomial coefficients using the Gamma function.

= Σ([(10C0)(5^10)/n^10] * (2/n)^0) + [(10C1)(5^9)(k(2/n))/n^9] * (2/n)^1 + [(10C2)(5^8)(k(2/n))^2/n^8] * (2/n)^2 + ...
= Σ([10!/0!(10-0)!)(5^10)/(n^10)] * (2/n)^0) + [10!/1!(10-1)!)(5^9)(k(2/n))/n^9] * (2/n)^1 + [10!/2!(10-2)!)(5^8)(k(2/n))^2/n^8] * (2/n)^2 + ...
= Σ([10!/(n^10)]) * (2/n)^0 * [(5^10)] + [10!/(n^9)] * (2/n)^1 * [(5^9)(k)] + [10!/(2!*n^8)] * (2/n)^2 * [(5^8)(k^2)] + ...

Step 4: Rewrite the sum as a definite integral
Now, we can rewrite the summation as a definite integral by replacing the Sigma notation with the integral sign, and the k variable with x.

= ∫[0, ∞] [10!/(n^10)] * (2/n)^0 * [(5^10)] + [10!/(n^9)] * (2/n)^1 * [(5^9)(x)] + [10!/(2!*n^8)] * (2/n)^2 * [(5^8)(x^2)] + ...

This is the expression of the given summation as a definite integral with the limits from 0 to infinity.