Using Mathematical Induction to Prove a Summation Formula

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The discussion focuses on using mathematical induction to prove the summation formula for \( S_k = 5 \cdot 6 + 5 \cdot 6^2 + \ldots + 5 \cdot 6^k = 6(6^k - 1) \). The next step involves proving \( S_{k+1} \) by adding \( 5 \cdot 6^{k+1} \) to both sides of the equation. The right-hand side is manipulated to show that it can be expressed as \( 6(6^{k+1} - 1) \). Through algebraic manipulation, terms are combined and factored to arrive at the desired form. The proof successfully demonstrates the validity of the summation formula using induction.
ineedhelpnow
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$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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The second statement should be labeled as $S_{k+1}$ and you want the right side to be:

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

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.
 
i don't know how to get there
 
here's all I've done so far:

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

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

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

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

$6(6^{k+1}-1)$
 
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Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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