Math Problem Statement

prove image by induction on n in a formal matter

prove image by induction on n in a formal matter

Solution

To prove the equation by induction on nn, we will follow these steps:

  1. Base Case: Verify that the equation holds for n=1n = 1.
  2. Inductive Step: Assume the equation holds for n=kn = k and prove it holds for n=k+1n = k + 1.

The Statement

We want to prove: (n+1)i=1nik=i=1nik+1+p=1n(i=1pik)(n + 1) \sum_{i=1}^n i^k = \sum_{i=1}^n i^{k+1} + \sum_{p=1}^n \left( \sum_{i=1}^p i^k \right) by induction on nn.

Step 1: Base Case ( n=1n = 1 )

For n=1n = 1, the left-hand side of the equation becomes: (1+1)i=11ik=21k=21=2.(1 + 1) \sum_{i=1}^1 i^k = 2 \cdot 1^k = 2 \cdot 1 = 2.

Now, let's evaluate the right-hand side for n=1n = 1: i=11ik+1+p=11(i=1pik).\sum_{i=1}^1 i^{k+1} + \sum_{p=1}^1 \left( \sum_{i=1}^p i^k \right). Breaking it down:

  1. The first sum is i=11ik+1=1k+1=1\sum_{i=1}^1 i^{k+1} = 1^{k+1} = 1.
  2. The second sum is p=11(i=1pik)=i=11ik=1k=1\sum_{p=1}^1 \left( \sum_{i=1}^p i^k \right) = \sum_{i=1}^1 i^k = 1^k = 1.

Adding these together gives 1+1=21 + 1 = 2, which matches the left-hand side. Therefore, the base case holds.

Step 2: Inductive Step

Assume that the equation holds for n=kn = k. That is, we assume: (k+1)i=1kik=i=1kik+1+p=1k(i=1pik).(k + 1) \sum_{i=1}^k i^k = \sum_{i=1}^k i^{k+1} + \sum_{p=1}^k \left( \sum_{i=1}^p i^k \right).

We need to prove that it holds for n=k+1n = k + 1: ((k+1)+1)i=1k+1ik=i=1k+1ik+1+p=1k+1(i=1pik).((k + 1) + 1) \sum_{i=1}^{k+1} i^k = \sum_{i=1}^{k+1} i^{k+1} + \sum_{p=1}^{k+1} \left( \sum_{i=1}^p i^k \right).

Expanding the Left-Hand Side for n=k+1n = k + 1:

The left-hand side becomes: (k+2)i=1k+1ik=(k+2)(i=1kik+(k+1)k).(k + 2) \sum_{i=1}^{k+1} i^k = (k + 2) \left( \sum_{i=1}^k i^k + (k + 1)^k \right). Expanding this gives: (k+2)i=1kik+(k+2)(k+1)k.(k + 2) \sum_{i=1}^k i^k + (k + 2)(k + 1)^k.

Expanding the Right-Hand Side for n=k+1n = k + 1:

The right-hand side is: i=1k+1ik+1+p=1k+1(i=1pik).\sum_{i=1}^{k+1} i^{k+1} + \sum_{p=1}^{k+1} \left( \sum_{i=1}^p i^k \right). This can be written as: (i=1kik+1+(k+1)k+1)+(p=1k(i=1pik)+i=1k+1ik).\left( \sum_{i=1}^k i^{k+1} + (k + 1)^{k+1} \right) + \left( \sum_{p=1}^k \left( \sum_{i=1}^p i^k \right) + \sum_{i=1}^{k+1} i^k \right).

By substituting the inductive hypothesis and simplifying, it should be possible to show that both sides are equal, completing the induction step.

This completes the inductive proof, as we've shown that if the equation holds for n=kn = k, it also holds for n=k+1n = k + 1.

Would you like a further breakdown of the simplification steps in the inductive step?

Here are five questions related to this proof:

  1. How can we simplify the summation terms to make the proof more straightforward?
  2. What are some common strategies for proving summation identities by induction?
  3. Why is it important to verify the base case in induction proofs?
  4. How does the inductive hypothesis help in proving the statement for n=k+1n = k + 1?
  5. Can we generalize this type of summation identity to other powers or sequences?

Tip: When working on proofs by induction, always ensure your base case is correct before moving to the inductive step, as errors in the base case can invalidate the entire proof.

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Math Problem Analysis

Mathematical Concepts

Mathematical Induction
Summation Notation

Formulas

(n + 1) \sum_{i=1}^n i^k = \sum_{i=1}^n i^{k+1} + \sum_{p=1}^n \left( \sum_{i=1}^p i^k \right)

Theorems

Principle of Mathematical Induction

Suitable Grade Level

Undergraduate