Mixing
Can someone please point out which induction's step I did wrong?
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$$sum_k=1^n k(k+1)(k+2)...(k+p-1)= fracn(n+1)(n+2)...(n+p) p+1$$
induction by substitute n+1:
$$(n+p+1)!= (n+p)!+(p+1)(n+p)! $$
I couldnt figure out how to countinue after this step. Thanks
discrete-mathematics induction recursion
asked Jul 21 at 9:23
stevie lol
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up vote
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$$sum_k=1^n k(k+1)(k+2)...(k+p-1)= fracn(n+1)(n+2)...(n+p) p+1$$
induction by substitute n+1:
$$(n+p+1)!= (n+p)!+(p+1)(n+p)! $$
I couldnt figure out how to countinue after this step. Thanks
discrete-mathematics induction recursion
asked Jul 21 at 9:23
stevie lol
306
2
If you want to decompose as a sum, it should be $;(n+p+1)!=n,(n+p)!+(p+1),(n+p)!$.
– Bernard
Jul 21 at 9:29
BTW, by dividing by $p!$, this is equivalent to $sum_k=1^n binomk+p-1p = binomn+pp+1$.
– lhf
Jul 21 at 10:09
1
No need of factorials. At the induction step you should show that $$(n+1)(n+2)cdots(n+p)= frac(n+1)cdots(n+p)(n+1+p) p+1-fracn(n+1)cdots(n+p) p+1.$$
– Robert Z
Jul 21 at 10:16
@Bernard how did u get n(n+p)! ?
– stevie lol
Jul 21 at 13:22
I simply wrote $(n+p+1)!=(n+p+1),(n+p)!$, split the first factor in two and used distributivity, nothing more.
– Bernard
Jul 21 at 13:25
 |Â
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$$sum_k=1^n k(k+1)(k+2)...(k+p-1)= fracn(n+1)(n+2)...(n+p) p+1$$
induction by substitute n+1:
$$(n+p+1)!= (n+p)!+(p+1)(n+p)! $$
I couldnt figure out how to countinue after this step. Thanks
discrete-mathematics induction recursion
asked Jul 21 at 9:23
stevie lol
306
$$sum_k=1^n k(k+1)(k+2)...(k+p-1)= fracn(n+1)(n+2)...(n+p) p+1$$
induction by substitute n+1:
$$(n+p+1)!= (n+p)!+(p+1)(n+p)! $$
I couldnt figure out how to countinue after this step. Thanks
discrete-mathematics induction recursion
asked Jul 21 at 9:23
stevie lol
306
asked Jul 21 at 9:23
stevie lol
306
asked Jul 21 at 9:23
stevie lol
306
asked Jul 21 at 9:23
stevie lol
306
306
2
If you want to decompose as a sum, it should be $;(n+p+1)!=n,(n+p)!+(p+1),(n+p)!$.
– Bernard
Jul 21 at 9:29
BTW, by dividing by $p!$, this is equivalent to $sum_k=1^n binomk+p-1p = binomn+pp+1$.
– lhf
Jul 21 at 10:09
1
No need of factorials. At the induction step you should show that $$(n+1)(n+2)cdots(n+p)= frac(n+1)cdots(n+p)(n+1+p) p+1-fracn(n+1)cdots(n+p) p+1.$$
– Robert Z
Jul 21 at 10:16
@Bernard how did u get n(n+p)! ?
– stevie lol
Jul 21 at 13:22
I simply wrote $(n+p+1)!=(n+p+1),(n+p)!$, split the first factor in two and used distributivity, nothing more.
– Bernard
Jul 21 at 13:25
 |Â
show 3 more comments
2
If you want to decompose as a sum, it should be $;(n+p+1)!=n,(n+p)!+(p+1),(n+p)!$.
– Bernard
Jul 21 at 9:29
BTW, by dividing by $p!$, this is equivalent to $sum_k=1^n binomk+p-1p = binomn+pp+1$.
– lhf
Jul 21 at 10:09
1
No need of factorials. At the induction step you should show that $$(n+1)(n+2)cdots(n+p)= frac(n+1)cdots(n+p)(n+1+p) p+1-fracn(n+1)cdots(n+p) p+1.$$
– Robert Z
Jul 21 at 10:16
@Bernard how did u get n(n+p)! ?
– stevie lol
Jul 21 at 13:22
I simply wrote $(n+p+1)!=(n+p+1),(n+p)!$, split the first factor in two and used distributivity, nothing more.
– Bernard
Jul 21 at 13:25
2
2
If you want to decompose as a sum, it should be $;(n+p+1)!=n,(n+p)!+(p+1),(n+p)!$.
– Bernard
Jul 21 at 9:29
If you want to decompose as a sum, it should be $;(n+p+1)!=n,(n+p)!+(p+1),(n+p)!$.
– Bernard
Jul 21 at 9:29
BTW, by dividing by $p!$, this is equivalent to $sum_k=1^n binomk+p-1p = binomn+pp+1$.
– lhf
Jul 21 at 10:09
BTW, by dividing by $p!$, this is equivalent to $sum_k=1^n binomk+p-1p = binomn+pp+1$.
– lhf
Jul 21 at 10:09
1
1
No need of factorials. At the induction step you should show that $$(n+1)(n+2)cdots(n+p)= frac(n+1)cdots(n+p)(n+1+p) p+1-fracn(n+1)cdots(n+p) p+1.$$
– Robert Z
Jul 21 at 10:16
No need of factorials. At the induction step you should show that $$(n+1)(n+2)cdots(n+p)= frac(n+1)cdots(n+p)(n+1+p) p+1-fracn(n+1)cdots(n+p) p+1.$$
– Robert Z
Jul 21 at 10:16
@Bernard how did u get n(n+p)! ?
– stevie lol
Jul 21 at 13:22
@Bernard how did u get n(n+p)! ?
– stevie lol
Jul 21 at 13:22
I simply wrote $(n+p+1)!=(n+p+1),(n+p)!$, split the first factor in two and used distributivity, nothing more.
– Bernard
Jul 21 at 13:25
I simply wrote $(n+p+1)!=(n+p+1),(n+p)!$, split the first factor in two and used distributivity, nothing more.
– Bernard
Jul 21 at 13:25
 |Â
show 3 more comments
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
Induction step:
$$sum_k=1^n+1 k(k+1)(k+2)...(k+p-1)=\
sum_k=1^n k(k+1)(k+2)...(k+p-1)+(n+1)(n+2)cdots(n+p)=\
fracn(n+1)(n+2)...(n+p) p+1+(n+1)(n+2)cdots(n+p)=\
frac(n+1)(n+1+1)...(n+1+p-1)(n+1+p) p+1.$$
answered Jul 21 at 10:31
farruhota
13.7k2632
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up vote
2
down vote
Suppose you have the answer for $n$.
For $n+1$, the sum will be $sumlimits_k=1^n+1 k(k+1)...(k+p-1)$
That is $sumlimits_k=1^n k(k+1)...(k+p-1) + (n+1)(n+2)...(n+p)$
Then, by induction hypothesis, you have $fracn(n+1)...(n+p)p+1+(n+1)(n+2)...(n+p)=fracn(n+1)...(n+p) + (n+1)(n+2)...(n+p)(p+1)p+1=frac((n+1)(n+2)...(n+p))(n + p + 1)p+1$
and you have your answer. Of course, you should also write down the basis for the induction.
answered Jul 21 at 10:36
Henrique Lecco
211
I couldn't simplify it. The furthest I got is (n+p)!(p+1)+(n+p)!=(n+1+p)!, did I do sth wrong?
– stevie lol
Jul 21 at 16:42
It's not that you've done something wrong. But I don't think you're getting anywhere with that approach. No need for factorials, try to use the induction hypothesis!
– Henrique Lecco
Jul 22 at 11:49
add a comment |Â
up vote
1
down vote
The expression $$(n+p+1)!= (n+p)!+(p+1)(n+p)!$$
should have been $$(n+p+1)!= (n+p+1)(n+p)! =n(n+p)!+(p+1)(n+p)!$$
answered Jul 21 at 10:22
Mohammad Riazi-Kermani
27.5k41852
1
Can i ask how did u get n(n+p)! ?
– stevie lol
Jul 21 at 13:20
1
Distribute n+p+1=n+(p+1) over (n+p)!
– Mohammad Riazi-Kermani
Jul 21 at 13:50
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Induction step:
$$sum_k=1^n+1 k(k+1)(k+2)...(k+p-1)=\
sum_k=1^n k(k+1)(k+2)...(k+p-1)+(n+1)(n+2)cdots(n+p)=\
fracn(n+1)(n+2)...(n+p) p+1+(n+1)(n+2)cdots(n+p)=\
frac(n+1)(n+1+1)...(n+1+p-1)(n+1+p) p+1.$$
answered Jul 21 at 10:31
farruhota
13.7k2632
add a comment |Â
up vote
2
down vote
accepted
Induction step:
$$sum_k=1^n+1 k(k+1)(k+2)...(k+p-1)=\
sum_k=1^n k(k+1)(k+2)...(k+p-1)+(n+1)(n+2)cdots(n+p)=\
fracn(n+1)(n+2)...(n+p) p+1+(n+1)(n+2)cdots(n+p)=\
frac(n+1)(n+1+1)...(n+1+p-1)(n+1+p) p+1.$$
answered Jul 21 at 10:31
farruhota
13.7k2632
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Induction step:
$$sum_k=1^n+1 k(k+1)(k+2)...(k+p-1)=\
sum_k=1^n k(k+1)(k+2)...(k+p-1)+(n+1)(n+2)cdots(n+p)=\
fracn(n+1)(n+2)...(n+p) p+1+(n+1)(n+2)cdots(n+p)=\
frac(n+1)(n+1+1)...(n+1+p-1)(n+1+p) p+1.$$
answered Jul 21 at 10:31
farruhota
13.7k2632
Induction step:
$$sum_k=1^n+1 k(k+1)(k+2)...(k+p-1)=\
sum_k=1^n k(k+1)(k+2)...(k+p-1)+(n+1)(n+2)cdots(n+p)=\
fracn(n+1)(n+2)...(n+p) p+1+(n+1)(n+2)cdots(n+p)=\
frac(n+1)(n+1+1)...(n+1+p-1)(n+1+p) p+1.$$
answered Jul 21 at 10:31
farruhota
13.7k2632
answered Jul 21 at 10:31
farruhota
13.7k2632
answered Jul 21 at 10:31
farruhota
13.7k2632
answered Jul 21 at 10:31
farruhota
13.7k2632
13.7k2632
add a comment |Â
add a comment |Â
up vote
2
down vote
Suppose you have the answer for $n$.
For $n+1$, the sum will be $sumlimits_k=1^n+1 k(k+1)...(k+p-1)$
That is $sumlimits_k=1^n k(k+1)...(k+p-1) + (n+1)(n+2)...(n+p)$
Then, by induction hypothesis, you have $fracn(n+1)...(n+p)p+1+(n+1)(n+2)...(n+p)=fracn(n+1)...(n+p) + (n+1)(n+2)...(n+p)(p+1)p+1=frac((n+1)(n+2)...(n+p))(n + p + 1)p+1$
and you have your answer. Of course, you should also write down the basis for the induction.
answered Jul 21 at 10:36
Henrique Lecco
211
I couldn't simplify it. The furthest I got is (n+p)!(p+1)+(n+p)!=(n+1+p)!, did I do sth wrong?
– stevie lol
Jul 21 at 16:42
It's not that you've done something wrong. But I don't think you're getting anywhere with that approach. No need for factorials, try to use the induction hypothesis!
– Henrique Lecco
Jul 22 at 11:49
add a comment |Â
up vote
2
down vote
Suppose you have the answer for $n$.
For $n+1$, the sum will be $sumlimits_k=1^n+1 k(k+1)...(k+p-1)$
That is $sumlimits_k=1^n k(k+1)...(k+p-1) + (n+1)(n+2)...(n+p)$
Then, by induction hypothesis, you have $fracn(n+1)...(n+p)p+1+(n+1)(n+2)...(n+p)=fracn(n+1)...(n+p) + (n+1)(n+2)...(n+p)(p+1)p+1=frac((n+1)(n+2)...(n+p))(n + p + 1)p+1$
and you have your answer. Of course, you should also write down the basis for the induction.
answered Jul 21 at 10:36
Henrique Lecco
211
I couldn't simplify it. The furthest I got is (n+p)!(p+1)+(n+p)!=(n+1+p)!, did I do sth wrong?
– stevie lol
Jul 21 at 16:42
It's not that you've done something wrong. But I don't think you're getting anywhere with that approach. No need for factorials, try to use the induction hypothesis!
– Henrique Lecco
Jul 22 at 11:49
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Suppose you have the answer for $n$.
For $n+1$, the sum will be $sumlimits_k=1^n+1 k(k+1)...(k+p-1)$
That is $sumlimits_k=1^n k(k+1)...(k+p-1) + (n+1)(n+2)...(n+p)$
Then, by induction hypothesis, you have $fracn(n+1)...(n+p)p+1+(n+1)(n+2)...(n+p)=fracn(n+1)...(n+p) + (n+1)(n+2)...(n+p)(p+1)p+1=frac((n+1)(n+2)...(n+p))(n + p + 1)p+1$
and you have your answer. Of course, you should also write down the basis for the induction.
answered Jul 21 at 10:36
Henrique Lecco
211
Suppose you have the answer for $n$.
For $n+1$, the sum will be $sumlimits_k=1^n+1 k(k+1)...(k+p-1)$
That is $sumlimits_k=1^n k(k+1)...(k+p-1) + (n+1)(n+2)...(n+p)$
Then, by induction hypothesis, you have $fracn(n+1)...(n+p)p+1+(n+1)(n+2)...(n+p)=fracn(n+1)...(n+p) + (n+1)(n+2)...(n+p)(p+1)p+1=frac((n+1)(n+2)...(n+p))(n + p + 1)p+1$
and you have your answer. Of course, you should also write down the basis for the induction.
answered Jul 21 at 10:36
Henrique Lecco
211
answered Jul 21 at 10:36
Henrique Lecco
211
answered Jul 21 at 10:36
Henrique Lecco
211
answered Jul 21 at 10:36
Henrique Lecco
211
211
I couldn't simplify it. The furthest I got is (n+p)!(p+1)+(n+p)!=(n+1+p)!, did I do sth wrong?
– stevie lol
Jul 21 at 16:42
It's not that you've done something wrong. But I don't think you're getting anywhere with that approach. No need for factorials, try to use the induction hypothesis!
– Henrique Lecco
Jul 22 at 11:49
add a comment |Â
I couldn't simplify it. The furthest I got is (n+p)!(p+1)+(n+p)!=(n+1+p)!, did I do sth wrong?
– stevie lol
Jul 21 at 16:42
It's not that you've done something wrong. But I don't think you're getting anywhere with that approach. No need for factorials, try to use the induction hypothesis!
– Henrique Lecco
Jul 22 at 11:49
I couldn't simplify it. The furthest I got is (n+p)!(p+1)+(n+p)!=(n+1+p)!, did I do sth wrong?
– stevie lol
Jul 21 at 16:42
I couldn't simplify it. The furthest I got is (n+p)!(p+1)+(n+p)!=(n+1+p)!, did I do sth wrong?
– stevie lol
Jul 21 at 16:42
It's not that you've done something wrong. But I don't think you're getting anywhere with that approach. No need for factorials, try to use the induction hypothesis!
– Henrique Lecco
Jul 22 at 11:49
It's not that you've done something wrong. But I don't think you're getting anywhere with that approach. No need for factorials, try to use the induction hypothesis!
– Henrique Lecco
Jul 22 at 11:49
add a comment |Â
up vote
1
down vote
The expression $$(n+p+1)!= (n+p)!+(p+1)(n+p)!$$
should have been $$(n+p+1)!= (n+p+1)(n+p)! =n(n+p)!+(p+1)(n+p)!$$
answered Jul 21 at 10:22
Mohammad Riazi-Kermani
27.5k41852
1
Can i ask how did u get n(n+p)! ?
– stevie lol
Jul 21 at 13:20
1
Distribute n+p+1=n+(p+1) over (n+p)!
– Mohammad Riazi-Kermani
Jul 21 at 13:50
add a comment |Â
up vote
1
down vote
The expression $$(n+p+1)!= (n+p)!+(p+1)(n+p)!$$
should have been $$(n+p+1)!= (n+p+1)(n+p)! =n(n+p)!+(p+1)(n+p)!$$
answered Jul 21 at 10:22
Mohammad Riazi-Kermani
27.5k41852
1
Can i ask how did u get n(n+p)! ?
– stevie lol
Jul 21 at 13:20
1
Distribute n+p+1=n+(p+1) over (n+p)!
– Mohammad Riazi-Kermani
Jul 21 at 13:50
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The expression $$(n+p+1)!= (n+p)!+(p+1)(n+p)!$$
should have been $$(n+p+1)!= (n+p+1)(n+p)! =n(n+p)!+(p+1)(n+p)!$$
answered Jul 21 at 10:22
Mohammad Riazi-Kermani
27.5k41852
The expression $$(n+p+1)!= (n+p)!+(p+1)(n+p)!$$
should have been $$(n+p+1)!= (n+p+1)(n+p)! =n(n+p)!+(p+1)(n+p)!$$
answered Jul 21 at 10:22
Mohammad Riazi-Kermani
27.5k41852
answered Jul 21 at 10:22
Mohammad Riazi-Kermani
27.5k41852
answered Jul 21 at 10:22
Mohammad Riazi-Kermani
27.5k41852
answered Jul 21 at 10:22
Mohammad Riazi-Kermani
27.5k41852
27.5k41852
1
Can i ask how did u get n(n+p)! ?
– stevie lol
Jul 21 at 13:20
1
Distribute n+p+1=n+(p+1) over (n+p)!
– Mohammad Riazi-Kermani
Jul 21 at 13:50
add a comment |Â
1
Can i ask how did u get n(n+p)! ?
– stevie lol
Jul 21 at 13:20
1
Distribute n+p+1=n+(p+1) over (n+p)!
– Mohammad Riazi-Kermani
Jul 21 at 13:50
1
1
Can i ask how did u get n(n+p)! ?
– stevie lol
Jul 21 at 13:20
Can i ask how did u get n(n+p)! ?
– stevie lol
Jul 21 at 13:20
1
1
Distribute n+p+1=n+(p+1) over (n+p)!
– Mohammad Riazi-Kermani
Jul 21 at 13:50
Distribute n+p+1=n+(p+1) over (n+p)!
– Mohammad Riazi-Kermani
Jul 21 at 13:50
add a comment |Â
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2
If you want to decompose as a sum, it should be $;(n+p+1)!=n,(n+p)!+(p+1),(n+p)!$.
– Bernard
Jul 21 at 9:29
BTW, by dividing by $p!$, this is equivalent to $sum_k=1^n binomk+p-1p = binomn+pp+1$.
– lhf
Jul 21 at 10:09
1
No need of factorials. At the induction step you should show that $$(n+1)(n+2)cdots(n+p)= frac(n+1)cdots(n+p)(n+1+p) p+1-fracn(n+1)cdots(n+p) p+1.$$
– Robert Z
Jul 21 at 10:16
@Bernard how did u get n(n+p)! ?
– stevie lol
Jul 21 at 13:22
I simply wrote $(n+p+1)!=(n+p+1),(n+p)!$, split the first factor in two and used distributivity, nothing more.
– Bernard
Jul 21 at 13:25