Math Problem Statement
∑ k=0 3k (n k ) = 4n TRY WITH INDUCTION
Solution
Let's prove the given summation identity using mathematical induction:
The identity to prove:
Base Case:
For , we check both sides of the equation.
On the right-hand side: So, the base case holds true, as both sides equal 1.
Inductive Hypothesis:
Assume that the formula holds for some arbitrary . That is, assume:
Inductive Step:
Now, we need to prove that the formula holds for , i.e., we need to show:
We can break the summation into two parts:
From Pascal's identity, we know:
Thus, we can rewrite the sum as:
Now split the sum into two parts:
The second sum can be rewritten by shifting the index of summation:
So, the two sums combined give us:
Substituting our inductive assumption: By the inductive hypothesis, we already know:
Thus, we now have:
Which is exactly what we wanted to prove.
Conclusion:
By induction, we have shown that the given identity holds true for all non-negative integers .
Would you like further clarification or details on any part of this proof?
Here are some related questions to explore:
- How does Pascal's identity relate to binomial expansions?
- What is the binomial theorem and how does it connect to this proof?
- Can this method of induction be used for other summation formulas?
- How can we prove other series summations using induction?
- What is the significance of the identity in combinatorics or other areas of mathematics?
Tip: Induction is a powerful tool, but always remember to carefully check the base case and ensure the inductive step is clear.
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Math Problem Analysis
Mathematical Concepts
Mathematical Induction
Binomial Coefficients
Summation Identities
Formulas
Sum of terms: ∑ (k=0 to n) 3^k (n choose k)
Binomial expansion: (1 + x)^n = ∑ (k=0 to n) (n choose k) x^k
Pascal's identity: (n+1 choose k) = (n choose k) + (n choose k-1)
Theorems
Principle of Mathematical Induction
Pascal's Identity
Suitable Grade Level
Grades 11-12 or early college
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