Using Mathematical Induction to Prove a Summation Formula

In summary, we can prove $S_{k+1}$ by combining like terms and factoring the right side of the equation $6(6^k-1)$.
  • #1
ineedhelpnow
651
0
$S_k:5\cdot 6 +5\cdot 6^2+5\cdot 6^3+ ...+5\cdot 6^k=6(6^k-1)$$S_k:5\cdot 6 +5\cdot 6^2+5\cdot 6^3+ ...+5\cdot 6^k+ 5\cdot 6^{k+1}=6(6^k-1)+5\cdot 6^{k+1}$

what do i do now? to prove $S_{k+1}$
 
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  • #2
The second statement should be labeled as $S_{k+1}$ and you want the right side to be:

\(\displaystyle 6\left(6^{k+1}-1\right)\)

Can you get from what you have to there algebraically?

By the way, I am going to move this thread to the Pre-Calculus subforum and retitle it to remove the abbreviation.
 
  • #3
i don't know how to get there
 
  • #4
here's all I've done so far:

RHS
$6^{k+1}-6+5\cdot 6^{k+1}$
 
  • #5
ineedhelpnow said:
here's all I've done so far:

RHS
$6^{k+1}-6+5\cdot 6^{k+1}$

Okay, combine like terms and then factor...
 
  • #6
$6^{k+1}-6+5\cdot 6^{k+1}$

$[6^{k+1}+5\cdot 6^{k+1}-6$

$[6^{k+1}(1+5)]-6$

$6[6^k(6)-1]$

$6(6^{k+1}-1)$
 
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FAQ: Using Mathematical Induction to Prove a Summation Formula

How does mathematical induction work?

Mathematical induction is a proof technique used to prove statements about all natural numbers. It involves two steps: the base case, where the statement is shown to be true for the first natural number, and the inductive step, where it is shown that if the statement is true for one natural number, it is also true for the next natural number. By repeating this process, the statement can be proven to be true for all natural numbers.

How is mathematical induction used to prove a summation formula?

In order to use mathematical induction to prove a summation formula, we must first define the formula and state the base case, which is usually when n=1. Then, we assume that the formula holds true for some value of n, and use this assumption to show that it also holds true for n+1. This completes the inductive step. By repeating this process, we can prove that the formula holds true for all natural numbers.

Can mathematical induction be used to prove any summation formula?

Yes, mathematical induction can be used to prove any summation formula, as long as the formula can be expressed as a sum of terms involving natural numbers. This is because mathematical induction is a proof technique that works for all natural numbers.

Are there any limitations to using mathematical induction to prove summation formulas?

One limitation of using mathematical induction to prove summation formulas is that it can only be used for formulas involving natural numbers. It cannot be used for formulas involving real or complex numbers. Additionally, the formula must be defined for all natural numbers, including 0.

How can we be sure that a summation formula proven using mathematical induction is correct?

Mathematical induction provides a rigorous method for proving statements about all natural numbers. By completing both the base case and inductive step, we can be confident that the summation formula holds true for all natural numbers. Additionally, it is always a good idea to double-check the proof and make sure all steps are correct.

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