Question

Use mathematical induction to prove that the statement is true for every positive integer n. 7*8+7*8^2+7*8^3...7*8^n=8(8^n-1)

Answer 1

7*8+7*8^2+7*8^3...7*8^n=8(8^n-1)

Answer 2

i think the sum is supposed to be 8^n - 1

Answer 3

You can write your sum in sigma notation as:
\[\large \sum_{i=1}^n 7\cdot8^i=7\sum_{i=1}^n 8^i=\]
1) Now: Base case: for n=1, LHS: = \(7\cdot8^1=56\), RHS = \(8(8^1-1)=8*7=56\)

2) Now, assume the formula is true for \(n=k\), that is
\[\large 7\sum_{i=1}^{k} 8^i=8(8^{k}-1)\]

3) Prove it's true for \(n=k+1\), that is
\[\large 7\sum_{i=1}^{k+1} 8^i=8(8^{k+1}-1)\]

Answer 4

then??

Answer 5

I dunno I'm trying to do it but it's not working somehow o.o

Answer 6

I also stop in this step..

Answer 7

Normally you would do this:
\[\large 7\sum_{i=1}^{k+1} 8^i=7\left( \sum_{i=1}^k8^i+8^{k+1}\right)\]
and then substitute \(\large \sum_{i=1}^k 8^i\) with \(\large 8(8^k-1)\) since that is your induction hypothesis.

Answer 8

and somehow show this would give in the end \(8(8^{k+1}-1)\)

Answer 9

Yah no I totally forgot about the 7: so this is what you do:

\[\large 7\sum_{i=1}^{k+1} 8^i=7\left( \sum_{i=1}^k8^i+8^{k+1}\right)=\underbrace{7\sum_{i=1}^k8^i}+7\cdot8^{k+1}\]
\[\large =\underbrace{8(8^k-1)}+7\cdot8^{k+1}\]\[\large =8^{k+1}-8+7\cdot 8^{k+1}\\\large =8\cdot 8^{k+1}-8\\ \large=8(8^{k+1}-1)\]

Answer 10

Since the formula works for n=k+1, then the formula works for all positive integer n.

Answer 11

yes!!!!!!!!! thank you so much!!!!!!!!!!!

Answer 12

:)