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Suppose we let $S_{p}=\sum_{k=1}^{n} k^{p} .$ When $p=1,$ we obtain $(2),$ when $p=2,$ we to obtain formulas for $S_{p}$ for positive integer values of $p$. We have already established the formula in the case $p=1,$ and now wish to find $S_{2}=\sum_{k=1}^{n} k^{2}$ we may proceed as follows:$$\begin{array}{c}(1+k)^{3}-k^{3}=1+3 k+3 k^{2} \\\sum_{k=1}^{n}\left((1+k)^{3}-k^{3}\right)= \\\sum_{k=1}^{n} 1+\sum_{k=1}^{n} 3 k+\sum_{k=1}^{n} 3 k^{2}= \\n+3 \sum_{k=1}^{n} k+n+3 \sum_{k=1}^{n} k^{2}=n+3 S_{1}+3 S_{2}\end{array}$$The first sum on the left is a telescoping sum with $a_{k}=k^{3},$ and by Exercise$20,$ it is easily determined. We know the value of $S_{1},$ so we may now solve for $S_{2} .$ Complete the details and find $S_{2}$.

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 5

Sigma Notation and Areas

Integrals

Oregon State University

Idaho State University

Boston College

Lectures

05:53

In mathematics, an indefinite integral is an integral whose integrand is not known in terms of elementary functions. An indefinite integral is usually encountered when integrating functions that are not elementary functions themselves.

40:35

In mathematics, integration is one of the two main operations of calculus, with its inverse operation, differentiation, being the other. Given a function of a real variable (often called "the integrand"), an antiderivative is a function whose derivative is the given function. The area under a real-valued function of a real variable is the integral of the function, provided it is defined on a closed interval around a given point. It is a basic result of calculus that an antiderivative always exists, and is equal to the original function evaluated at the upper limit of integration.

01:07

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Yes. So the formula can be derived as followed will begin with the formula Cape was m cubed minus K cubed, which equals three K squared plus three K plus one. And we're gonna write this formula for K equals one through N. Um So using this method we can find one plus 1 cubed plus all the way to end cube and we see that this is going to be similar to another formula that is shown or shown the steps before. So we can keep using these steps to get one to the fourth all the way to enter the fourth. This is ultimately what we want to show. Uh and we'll see that it ends up giving us a final formula that's similar to the form Sean, but below.

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