How do i do this mathematical induction question?
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My question:$5+10+20+...+5(2)^{n-1} = 5(2^n -1)$
- So first step i have to prove LHS = RHS when $n=1$, which is true.
- Then I assume the statement is true for $n=k$
- Since the statement is true for $n=k$ then for $n=k+1$
My workings:
$5+10+20+...+5(2)^{k-1} +5(2)^{(k+1)-1}= 5(2^{k+1} -1)$
LHS: $5(2^{k-1}) + 5(2)^k$
Then I do not know how to proceed to simplify, in general, can someone show some steps and show me how to tackle simplifying this kind of questions?
algebra-precalculus induction exponential-function
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up vote
0
down vote
favorite
My question:$5+10+20+...+5(2)^{n-1} = 5(2^n -1)$
- So first step i have to prove LHS = RHS when $n=1$, which is true.
- Then I assume the statement is true for $n=k$
- Since the statement is true for $n=k$ then for $n=k+1$
My workings:
$5+10+20+...+5(2)^{k-1} +5(2)^{(k+1)-1}= 5(2^{k+1} -1)$
LHS: $5(2^{k-1}) + 5(2)^k$
Then I do not know how to proceed to simplify, in general, can someone show some steps and show me how to tackle simplifying this kind of questions?
algebra-precalculus induction exponential-function
LHS is $5(2^k-1)+5 cdot2^k$!
– Jens Schwaiger
yesterday
I don't know how you came up with the term after 'LHS' - it seems to be wrong. After you use the induction hypothesis and replace the sum it should read $5(2^k-1)+5cdot 2^k$
– maxmilgram
yesterday
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
My question:$5+10+20+...+5(2)^{n-1} = 5(2^n -1)$
- So first step i have to prove LHS = RHS when $n=1$, which is true.
- Then I assume the statement is true for $n=k$
- Since the statement is true for $n=k$ then for $n=k+1$
My workings:
$5+10+20+...+5(2)^{k-1} +5(2)^{(k+1)-1}= 5(2^{k+1} -1)$
LHS: $5(2^{k-1}) + 5(2)^k$
Then I do not know how to proceed to simplify, in general, can someone show some steps and show me how to tackle simplifying this kind of questions?
algebra-precalculus induction exponential-function
My question:$5+10+20+...+5(2)^{n-1} = 5(2^n -1)$
- So first step i have to prove LHS = RHS when $n=1$, which is true.
- Then I assume the statement is true for $n=k$
- Since the statement is true for $n=k$ then for $n=k+1$
My workings:
$5+10+20+...+5(2)^{k-1} +5(2)^{(k+1)-1}= 5(2^{k+1} -1)$
LHS: $5(2^{k-1}) + 5(2)^k$
Then I do not know how to proceed to simplify, in general, can someone show some steps and show me how to tackle simplifying this kind of questions?
algebra-precalculus induction exponential-function
algebra-precalculus induction exponential-function
asked yesterday
Tfue
486
486
LHS is $5(2^k-1)+5 cdot2^k$!
– Jens Schwaiger
yesterday
I don't know how you came up with the term after 'LHS' - it seems to be wrong. After you use the induction hypothesis and replace the sum it should read $5(2^k-1)+5cdot 2^k$
– maxmilgram
yesterday
add a comment |
LHS is $5(2^k-1)+5 cdot2^k$!
– Jens Schwaiger
yesterday
I don't know how you came up with the term after 'LHS' - it seems to be wrong. After you use the induction hypothesis and replace the sum it should read $5(2^k-1)+5cdot 2^k$
– maxmilgram
yesterday
LHS is $5(2^k-1)+5 cdot2^k$!
– Jens Schwaiger
yesterday
LHS is $5(2^k-1)+5 cdot2^k$!
– Jens Schwaiger
yesterday
I don't know how you came up with the term after 'LHS' - it seems to be wrong. After you use the induction hypothesis and replace the sum it should read $5(2^k-1)+5cdot 2^k$
– maxmilgram
yesterday
I don't know how you came up with the term after 'LHS' - it seems to be wrong. After you use the induction hypothesis and replace the sum it should read $5(2^k-1)+5cdot 2^k$
– maxmilgram
yesterday
add a comment |
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
$5(2^k - 1) + 5(2^k) = 5(2^{k+1} -1)$
how? what did you do?
– Tfue
yesterday
yo all g, i just need to get me eyes check!
– Tfue
yesterday
add a comment |
up vote
0
down vote
For $n=1$ the case is true. So assume that the case for $n=k$ is also true i.e. $5+10+20+cdots +5(2)^{k}=5(2^k+1)$. So we shall prove for $n=k+1$ viz $5+10+20+cdots +5(2)^{k}+5(2)^{k+1}=5(2^k+1)+5(2)^{k+1}
=5(2^k+1+2^{k+1})=5(2^{k+2}-1)$
So the result is true for $n=k+1$. By the Principle of mathematical induction the result is true for all $n$.
add a comment |
up vote
0
down vote
Well, induction is not necessary.
The equation can be simplified by canceling $5$ from both sides.
We are left with
$$1+2+4..2^{n-1}=2^n -1$$
The LHS simply happens to be the sum of $n$ terms of a GP such that $a=1$ and $r=2$.
The sum $S_n$ is
$$S_n= frac{a(r^n -1)}{r-1}$$
$$=frac{(1)(2^n -1)}{2-1}$$
$$=2^n -1$$
Hence proved.
Hope this helps.
New contributor
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$5(2^k - 1) + 5(2^k) = 5(2^{k+1} -1)$
how? what did you do?
– Tfue
yesterday
yo all g, i just need to get me eyes check!
– Tfue
yesterday
add a comment |
up vote
1
down vote
accepted
$5(2^k - 1) + 5(2^k) = 5(2^{k+1} -1)$
how? what did you do?
– Tfue
yesterday
yo all g, i just need to get me eyes check!
– Tfue
yesterday
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$5(2^k - 1) + 5(2^k) = 5(2^{k+1} -1)$
$5(2^k - 1) + 5(2^k) = 5(2^{k+1} -1)$
answered yesterday
MoonKnight
1,189510
1,189510
how? what did you do?
– Tfue
yesterday
yo all g, i just need to get me eyes check!
– Tfue
yesterday
add a comment |
how? what did you do?
– Tfue
yesterday
yo all g, i just need to get me eyes check!
– Tfue
yesterday
how? what did you do?
– Tfue
yesterday
how? what did you do?
– Tfue
yesterday
yo all g, i just need to get me eyes check!
– Tfue
yesterday
yo all g, i just need to get me eyes check!
– Tfue
yesterday
add a comment |
up vote
0
down vote
For $n=1$ the case is true. So assume that the case for $n=k$ is also true i.e. $5+10+20+cdots +5(2)^{k}=5(2^k+1)$. So we shall prove for $n=k+1$ viz $5+10+20+cdots +5(2)^{k}+5(2)^{k+1}=5(2^k+1)+5(2)^{k+1}
=5(2^k+1+2^{k+1})=5(2^{k+2}-1)$
So the result is true for $n=k+1$. By the Principle of mathematical induction the result is true for all $n$.
add a comment |
up vote
0
down vote
For $n=1$ the case is true. So assume that the case for $n=k$ is also true i.e. $5+10+20+cdots +5(2)^{k}=5(2^k+1)$. So we shall prove for $n=k+1$ viz $5+10+20+cdots +5(2)^{k}+5(2)^{k+1}=5(2^k+1)+5(2)^{k+1}
=5(2^k+1+2^{k+1})=5(2^{k+2}-1)$
So the result is true for $n=k+1$. By the Principle of mathematical induction the result is true for all $n$.
add a comment |
up vote
0
down vote
up vote
0
down vote
For $n=1$ the case is true. So assume that the case for $n=k$ is also true i.e. $5+10+20+cdots +5(2)^{k}=5(2^k+1)$. So we shall prove for $n=k+1$ viz $5+10+20+cdots +5(2)^{k}+5(2)^{k+1}=5(2^k+1)+5(2)^{k+1}
=5(2^k+1+2^{k+1})=5(2^{k+2}-1)$
So the result is true for $n=k+1$. By the Principle of mathematical induction the result is true for all $n$.
For $n=1$ the case is true. So assume that the case for $n=k$ is also true i.e. $5+10+20+cdots +5(2)^{k}=5(2^k+1)$. So we shall prove for $n=k+1$ viz $5+10+20+cdots +5(2)^{k}+5(2)^{k+1}=5(2^k+1)+5(2)^{k+1}
=5(2^k+1+2^{k+1})=5(2^{k+2}-1)$
So the result is true for $n=k+1$. By the Principle of mathematical induction the result is true for all $n$.
answered yesterday
Yadati Kiran
231110
231110
add a comment |
add a comment |
up vote
0
down vote
Well, induction is not necessary.
The equation can be simplified by canceling $5$ from both sides.
We are left with
$$1+2+4..2^{n-1}=2^n -1$$
The LHS simply happens to be the sum of $n$ terms of a GP such that $a=1$ and $r=2$.
The sum $S_n$ is
$$S_n= frac{a(r^n -1)}{r-1}$$
$$=frac{(1)(2^n -1)}{2-1}$$
$$=2^n -1$$
Hence proved.
Hope this helps.
New contributor
add a comment |
up vote
0
down vote
Well, induction is not necessary.
The equation can be simplified by canceling $5$ from both sides.
We are left with
$$1+2+4..2^{n-1}=2^n -1$$
The LHS simply happens to be the sum of $n$ terms of a GP such that $a=1$ and $r=2$.
The sum $S_n$ is
$$S_n= frac{a(r^n -1)}{r-1}$$
$$=frac{(1)(2^n -1)}{2-1}$$
$$=2^n -1$$
Hence proved.
Hope this helps.
New contributor
add a comment |
up vote
0
down vote
up vote
0
down vote
Well, induction is not necessary.
The equation can be simplified by canceling $5$ from both sides.
We are left with
$$1+2+4..2^{n-1}=2^n -1$$
The LHS simply happens to be the sum of $n$ terms of a GP such that $a=1$ and $r=2$.
The sum $S_n$ is
$$S_n= frac{a(r^n -1)}{r-1}$$
$$=frac{(1)(2^n -1)}{2-1}$$
$$=2^n -1$$
Hence proved.
Hope this helps.
New contributor
Well, induction is not necessary.
The equation can be simplified by canceling $5$ from both sides.
We are left with
$$1+2+4..2^{n-1}=2^n -1$$
The LHS simply happens to be the sum of $n$ terms of a GP such that $a=1$ and $r=2$.
The sum $S_n$ is
$$S_n= frac{a(r^n -1)}{r-1}$$
$$=frac{(1)(2^n -1)}{2-1}$$
$$=2^n -1$$
Hence proved.
Hope this helps.
New contributor
New contributor
answered 45 mins ago
AryanSonwatikar
195
195
New contributor
New contributor
add a comment |
add a comment |
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LHS is $5(2^k-1)+5 cdot2^k$!
– Jens Schwaiger
yesterday
I don't know how you came up with the term after 'LHS' - it seems to be wrong. After you use the induction hypothesis and replace the sum it should read $5(2^k-1)+5cdot 2^k$
– maxmilgram
yesterday