Prove part (c) of Theorem 7.4 .2 . [Hint: Begin with the difference (k+1)^4-k^4 and follow the s

Prove part (c) of Theorem 7.4 .2 . [Hint: Begin with the...

Key Concepts

Mathematical Induction

Mathematical induction is a proof technique that demonstrates the truth of an infinite sequence of propositions. It involves proving a base case, and then showing that if the statement holds for an arbitrary case k, it also holds for the case k+1. This method is especially powerful for proofs involving sums and recursive formulas, and it establishes rigorous results step by step.

Telescoping Series

A telescoping series is a sum where many terms cancel out when consecutive differences are taken, leaving only a few terms from the beginning and end of the sequence. By writing the difference between consecutive terms of a sequence, one can simplify and evaluate sums that might otherwise seem complex. This principle is useful to sum expressions that arise from the difference of powers.

Binomial Expansion

The binomial expansion is a method used to expand expressions raised to a power, particularly in the form (k+1)^n. By expanding such an expression into a sum of powers and coefficients, one obtains a formula that reveals the structure of successive differences, which forms the basis for deriving results in summation formulas and proofs involving powers.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying expressions to reveal underlying patterns or to facilitate further analysis. Techniques such as factoring, expanding, and simplifying polynomial expressions are essential for transforming complex differences into a more manageable form, which is a common strategy in proofs involving series and sequences.

Transcript

00:01 For this part, or for this problem, we are asked to prove part c of theorem 7 .4 .2, that being that the sum from k equals 1 up to n of k cubed equals n times n plus 1, oops, n times n plus 1, all over 2, all squared.

00:21 So, what we can do is first note that the sum from k equals 1 up to n of k plus 1 to the power of 4 minus k to the power of 4 would equal, if we expand out those ks, would be equal to the sum from k equals 1 up to n of 4 times k cubed plus 6 times k squared plus 4 times k squared plus 4 times k, plus 1.

00:55 So that means then that we can express k cubed as being equal to 1 over 4 times the sum from k equals 1 up to n of k plus 1 to the power of 4.

01:17 Oops, k plus 1, power of 4 minus k to the power of 4.

01:21 Oops, k to the power of 4.

01:24 That's our first sum and then we can write down minus then 6k squared or minus the sum from k equals 1 up to n of 6k squared minus the sum from k equals 1 up to n of 4k and then minus the sum from k equals 1 to n of just 1 so let's put some special attention onto those square onto that square bracket sum, because the other summations there we can figure out using the previous parts, you know, parts a and b of theorem 7 .4 .2.

02:06 So we can write out that sum.

02:07 If we write out the first few terms of it, well, it would be two to the power of four minus one to the power of four, then plus three to the power of four, minus 2 to the power of 4.

02:22 So we can see that we have two terms adding up to 0 there.

02:25 Then we would have plus 4 to the power of 4 minus 3 to the power of 4.

02:30 So we have another cancellation.

02:32 And then we would have plus 5 to the power of 4 minus 4 to the power of 4...

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