Proof by Mathematical Induction

1) Does it work for ANYTHING? Demonstrate. Try n = 1.
2) Given that it works for something, does it work for the NEXT thing? Prove it.
 
The reason you see so many sums in "induction" problems is that \(\displaystyle \sum_{r=1}^{k+1} a_r= \sum_{r=1}^k a_r+ a_{r+1}\).

IF it is true that \(\displaystyle \sum_{r= 1}^k (6r+ 5)= k(3k+ 8)\) then \(\displaystyle \sum_{r=1}^{k+1}= k(3k+ 8)+ 6(k+1)+ 5.\)
 
IF it is true that \(\displaystyle \sum_{r= 1}^k (6r+ 5)= k(3k+ 8)\), then \(\displaystyle \sum_{r=1}^{k+1}(6r + 5) = k(3k+ 8)+ 6(k+1)+ 5.\)

 
Jomo said:
The term broken off the sum is wrong. It literally is the last term on the 1st line.
Here is the line \(\displaystyle \sum_{r=1}^{k+1} a_r= \sum_{r=1}^k a_r+ a_{r+1}\)
Click to expand...
Ah. [math]\sum_{r = 1}^{k + 1} a_r = \sum_{r = 1}^k a_r + a_{k + 1}[/math].

I see now. Thanks!

-Dan
 
mathboyyy said:
Hi. Can someone help me with this please?View attachment 14995View attachment 14995
Click to expand...
Suppose that \(\displaystyle \sum\limits_{r - 1}^K {(6r + 5)} = K(3K + 8)\) is known to be true.
We now consider \(\displaystyle \begin{gathered}\\
\sum\limits_{r - 1}^{K + 1} {(6r + 5)} = \sum\limits_{r - 1}^K {(6r + 5)} + \left[ {6(K + 1) + 5} \right] \\
= K(3K + 8) + \left[ {6K + 11} \right] \\
= 3{K^2} + 14k + 11 \\
= (K + 1)(3K + 11) \\
= (K + 1)\left[ {3(K + 1) + 8} \right] \\
\end{gathered}\ \)
 
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