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What does p stands for and how do i use induction to solve it?
Clash Royale CLAN TAG#URR8PPP
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$$sum_k=1^n k(k+1)(k+2)...(k+p-1)= fracn(n+1)(n+2)...(n+p) p+1 $$
Hi, can someone explain what does p means in the equation?
I know I have to substitute 1 as a base case, but I don't know where, and I also know I also have to substitute n+1 for the induction part, but I don't know how? thanks
induction recursion
asked Jul 18 at 15:47
stevie lol
306
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up vote
0
down vote
favorite
$$sum_k=1^n k(k+1)(k+2)...(k+p-1)= fracn(n+1)(n+2)...(n+p) p+1 $$
Hi, can someone explain what does p means in the equation?
I know I have to substitute 1 as a base case, but I don't know where, and I also know I also have to substitute n+1 for the induction part, but I don't know how? thanks
induction recursion
asked Jul 18 at 15:47
stevie lol
306
2
$p$ is just an arbitrary positive integer.
– Lord Shark the Unknown
Jul 18 at 15:49
By the looks of it, $p$ is an arbitrary natural number.
– Clive Newstead
Jul 18 at 15:49
I would assume it is an arbitrary, yet constant, natural numbers, so that $k(k+1)....(k+ p -1)$ is the product of $p$ consecutive number. So the sum is $1*2*...*p + 2*3*4*....*(p+1) + 3*4*5*.....*(p+3)+ ...... + n*(n+1)....(n+p-1)$.
– fleablood
Jul 18 at 15:53
1
I think you need to do induction on $n$, and $p$ is just an arbitrary number, which can be anything. So just leave it as $p$. Unless you're finding that difficult, in which case, you could try proving by induction on $n$ for fixed $p$ to see what's going on, before you attempt a proof for arbitrary $p$.
– Antinous
Jul 18 at 15:54
2
okay factorials. $sum_i=1^n frac (n+ p-1)!(n-1)! = frac (n+p)!(n-1)!(p+1)$ is what needs to be proven. So $sum_i=1^n+1 frac (n+ p-1)!(n-1)! = [sum_i=1^n frac (n+ p-1)!(n-1)!] + frac (n+p)!n! = frac (n+p)!(n-1)!(p+1) + frac (n+p)!n!=frac (n+p)!(n-1)!(p+1)*frac nn + frac (n+p)!n!frac p+1p+1 = frac (n+p)!n!(p+1)[n + (p+1)] = frac (n+p+1)!n!(p+1)$. That's it.
– fleablood
Jul 21 at 19:18
 |Â
show 5 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$$sum_k=1^n k(k+1)(k+2)...(k+p-1)= fracn(n+1)(n+2)...(n+p) p+1 $$
Hi, can someone explain what does p means in the equation?
I know I have to substitute 1 as a base case, but I don't know where, and I also know I also have to substitute n+1 for the induction part, but I don't know how? thanks
induction recursion
asked Jul 18 at 15:47
stevie lol
306
$$sum_k=1^n k(k+1)(k+2)...(k+p-1)= fracn(n+1)(n+2)...(n+p) p+1 $$
Hi, can someone explain what does p means in the equation?
I know I have to substitute 1 as a base case, but I don't know where, and I also know I also have to substitute n+1 for the induction part, but I don't know how? thanks
induction recursion
asked Jul 18 at 15:47
stevie lol
306
asked Jul 18 at 15:47
stevie lol
306
asked Jul 18 at 15:47
stevie lol
306
asked Jul 18 at 15:47
stevie lol
306
306
2
$p$ is just an arbitrary positive integer.
– Lord Shark the Unknown
Jul 18 at 15:49
By the looks of it, $p$ is an arbitrary natural number.
– Clive Newstead
Jul 18 at 15:49
I would assume it is an arbitrary, yet constant, natural numbers, so that $k(k+1)....(k+ p -1)$ is the product of $p$ consecutive number. So the sum is $1*2*...*p + 2*3*4*....*(p+1) + 3*4*5*.....*(p+3)+ ...... + n*(n+1)....(n+p-1)$.
– fleablood
Jul 18 at 15:53
1
I think you need to do induction on $n$, and $p$ is just an arbitrary number, which can be anything. So just leave it as $p$. Unless you're finding that difficult, in which case, you could try proving by induction on $n$ for fixed $p$ to see what's going on, before you attempt a proof for arbitrary $p$.
– Antinous
Jul 18 at 15:54
2
okay factorials. $sum_i=1^n frac (n+ p-1)!(n-1)! = frac (n+p)!(n-1)!(p+1)$ is what needs to be proven. So $sum_i=1^n+1 frac (n+ p-1)!(n-1)! = [sum_i=1^n frac (n+ p-1)!(n-1)!] + frac (n+p)!n! = frac (n+p)!(n-1)!(p+1) + frac (n+p)!n!=frac (n+p)!(n-1)!(p+1)*frac nn + frac (n+p)!n!frac p+1p+1 = frac (n+p)!n!(p+1)[n + (p+1)] = frac (n+p+1)!n!(p+1)$. That's it.
– fleablood
Jul 21 at 19:18
 |Â
show 5 more comments
2
$p$ is just an arbitrary positive integer.
– Lord Shark the Unknown
Jul 18 at 15:49
By the looks of it, $p$ is an arbitrary natural number.
– Clive Newstead
Jul 18 at 15:49
I would assume it is an arbitrary, yet constant, natural numbers, so that $k(k+1)....(k+ p -1)$ is the product of $p$ consecutive number. So the sum is $1*2*...*p + 2*3*4*....*(p+1) + 3*4*5*.....*(p+3)+ ...... + n*(n+1)....(n+p-1)$.
– fleablood
Jul 18 at 15:53
1
I think you need to do induction on $n$, and $p$ is just an arbitrary number, which can be anything. So just leave it as $p$. Unless you're finding that difficult, in which case, you could try proving by induction on $n$ for fixed $p$ to see what's going on, before you attempt a proof for arbitrary $p$.
– Antinous
Jul 18 at 15:54
2
okay factorials. $sum_i=1^n frac (n+ p-1)!(n-1)! = frac (n+p)!(n-1)!(p+1)$ is what needs to be proven. So $sum_i=1^n+1 frac (n+ p-1)!(n-1)! = [sum_i=1^n frac (n+ p-1)!(n-1)!] + frac (n+p)!n! = frac (n+p)!(n-1)!(p+1) + frac (n+p)!n!=frac (n+p)!(n-1)!(p+1)*frac nn + frac (n+p)!n!frac p+1p+1 = frac (n+p)!n!(p+1)[n + (p+1)] = frac (n+p+1)!n!(p+1)$. That's it.
– fleablood
Jul 21 at 19:18
2
2
$p$ is just an arbitrary positive integer.
– Lord Shark the Unknown
Jul 18 at 15:49
$p$ is just an arbitrary positive integer.
– Lord Shark the Unknown
Jul 18 at 15:49
By the looks of it, $p$ is an arbitrary natural number.
– Clive Newstead
Jul 18 at 15:49
By the looks of it, $p$ is an arbitrary natural number.
– Clive Newstead
Jul 18 at 15:49
I would assume it is an arbitrary, yet constant, natural numbers, so that $k(k+1)....(k+ p -1)$ is the product of $p$ consecutive number. So the sum is $1*2*...*p + 2*3*4*....*(p+1) + 3*4*5*.....*(p+3)+ ...... + n*(n+1)....(n+p-1)$.
– fleablood
Jul 18 at 15:53
I would assume it is an arbitrary, yet constant, natural numbers, so that $k(k+1)....(k+ p -1)$ is the product of $p$ consecutive number. So the sum is $1*2*...*p + 2*3*4*....*(p+1) + 3*4*5*.....*(p+3)+ ...... + n*(n+1)....(n+p-1)$.
– fleablood
Jul 18 at 15:53
1
1
I think you need to do induction on $n$, and $p$ is just an arbitrary number, which can be anything. So just leave it as $p$. Unless you're finding that difficult, in which case, you could try proving by induction on $n$ for fixed $p$ to see what's going on, before you attempt a proof for arbitrary $p$.
– Antinous
Jul 18 at 15:54
I think you need to do induction on $n$, and $p$ is just an arbitrary number, which can be anything. So just leave it as $p$. Unless you're finding that difficult, in which case, you could try proving by induction on $n$ for fixed $p$ to see what's going on, before you attempt a proof for arbitrary $p$.
– Antinous
Jul 18 at 15:54
2
2
okay factorials. $sum_i=1^n frac (n+ p-1)!(n-1)! = frac (n+p)!(n-1)!(p+1)$ is what needs to be proven. So $sum_i=1^n+1 frac (n+ p-1)!(n-1)! = [sum_i=1^n frac (n+ p-1)!(n-1)!] + frac (n+p)!n! = frac (n+p)!(n-1)!(p+1) + frac (n+p)!n!=frac (n+p)!(n-1)!(p+1)*frac nn + frac (n+p)!n!frac p+1p+1 = frac (n+p)!n!(p+1)[n + (p+1)] = frac (n+p+1)!n!(p+1)$. That's it.
– fleablood
Jul 21 at 19:18
okay factorials. $sum_i=1^n frac (n+ p-1)!(n-1)! = frac (n+p)!(n-1)!(p+1)$ is what needs to be proven. So $sum_i=1^n+1 frac (n+ p-1)!(n-1)! = [sum_i=1^n frac (n+ p-1)!(n-1)!] + frac (n+p)!n! = frac (n+p)!(n-1)!(p+1) + frac (n+p)!n!=frac (n+p)!(n-1)!(p+1)*frac nn + frac (n+p)!n!frac p+1p+1 = frac (n+p)!n!(p+1)[n + (p+1)] = frac (n+p+1)!n!(p+1)$. That's it.
– fleablood
Jul 21 at 19:18
 |Â
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2
$p$ is just an arbitrary positive integer.
– Lord Shark the Unknown
Jul 18 at 15:49
By the looks of it, $p$ is an arbitrary natural number.
– Clive Newstead
Jul 18 at 15:49
I would assume it is an arbitrary, yet constant, natural numbers, so that $k(k+1)....(k+ p -1)$ is the product of $p$ consecutive number. So the sum is $1*2*...*p + 2*3*4*....*(p+1) + 3*4*5*.....*(p+3)+ ...... + n*(n+1)....(n+p-1)$.
– fleablood
Jul 18 at 15:53
1
I think you need to do induction on $n$, and $p$ is just an arbitrary number, which can be anything. So just leave it as $p$. Unless you're finding that difficult, in which case, you could try proving by induction on $n$ for fixed $p$ to see what's going on, before you attempt a proof for arbitrary $p$.
– Antinous
Jul 18 at 15:54
2
okay factorials. $sum_i=1^n frac (n+ p-1)!(n-1)! = frac (n+p)!(n-1)!(p+1)$ is what needs to be proven. So $sum_i=1^n+1 frac (n+ p-1)!(n-1)! = [sum_i=1^n frac (n+ p-1)!(n-1)!] + frac (n+p)!n! = frac (n+p)!(n-1)!(p+1) + frac (n+p)!n!=frac (n+p)!(n-1)!(p+1)*frac nn + frac (n+p)!n!frac p+1p+1 = frac (n+p)!n!(p+1)[n + (p+1)] = frac (n+p+1)!n!(p+1)$. That's it.
– fleablood
Jul 21 at 19:18