A question on binomial theorem
$begingroup$
If $C_0$, $C_1$, $C_2$,...$C_n$ are the coefficients in the expansion of $(1+x)^n$, where $n$ is a positive integer, show that $$C_1- {C_2over 2}
+{C_3over 3}-...+{(-1)^{n-1} C_nover n}=1+ {1over 2}+ {1over 3}+...+{1over n}$$
binomial-coefficients binomial-theorem
edited Jan 10 at 12:15
darij grinberg
11.5k33168
asked Jan 10 at 12:14
Supriyo HalderSupriyo Halder
658113
$endgroup$
add a comment |
$begingroup$
If $C_0$, $C_1$, $C_2$,...$C_n$ are the coefficients in the expansion of $(1+x)^n$, where $n$ is a positive integer, show that $$C_1- {C_2over 2}
+{C_3over 3}-...+{(-1)^{n-1} C_nover n}=1+ {1over 2}+ {1over 3}+...+{1over n}$$
binomial-coefficients binomial-theorem
edited Jan 10 at 12:15
darij grinberg
11.5k33168
asked Jan 10 at 12:14
Supriyo HalderSupriyo Halder
658113
$endgroup$
$begingroup$
math.stackexchange.com/questions/437523/…
$endgroup$
– lab bhattacharjee
Jan 10 at 12:29
add a comment |
$begingroup$
If $C_0$, $C_1$, $C_2$,...$C_n$ are the coefficients in the expansion of $(1+x)^n$, where $n$ is a positive integer, show that $$C_1- {C_2over 2}
+{C_3over 3}-...+{(-1)^{n-1} C_nover n}=1+ {1over 2}+ {1over 3}+...+{1over n}$$
binomial-coefficients binomial-theorem
edited Jan 10 at 12:15
darij grinberg
11.5k33168
asked Jan 10 at 12:14
Supriyo HalderSupriyo Halder
658113
$endgroup$
If $C_0$, $C_1$, $C_2$,...$C_n$ are the coefficients in the expansion of $(1+x)^n$, where $n$ is a positive integer, show that $$C_1- {C_2over 2}
+{C_3over 3}-...+{(-1)^{n-1} C_nover n}=1+ {1over 2}+ {1over 3}+...+{1over n}$$
binomial-coefficients binomial-theorem
binomial-coefficients binomial-theorem
edited Jan 10 at 12:15
darij grinberg
11.5k33168
asked Jan 10 at 12:14
Supriyo HalderSupriyo Halder
658113
edited Jan 10 at 12:15
darij grinberg
11.5k33168
asked Jan 10 at 12:14
Supriyo HalderSupriyo Halder
658113
edited Jan 10 at 12:15
darij grinberg
11.5k33168
edited Jan 10 at 12:15
darij grinberg
11.5k33168
edited Jan 10 at 12:15
darij grinberg
11.5k33168
11.5k33168
asked Jan 10 at 12:14
Supriyo HalderSupriyo Halder
658113
asked Jan 10 at 12:14
Supriyo HalderSupriyo Halder
658113
asked Jan 10 at 12:14
Supriyo HalderSupriyo Halder
658113
658113
$begingroup$
math.stackexchange.com/questions/437523/…
$endgroup$
– lab bhattacharjee
Jan 10 at 12:29
add a comment |
$begingroup$
math.stackexchange.com/questions/437523/…
$endgroup$
– lab bhattacharjee
Jan 10 at 12:29
$begingroup$
math.stackexchange.com/questions/437523/…
$endgroup$
– lab bhattacharjee
Jan 10 at 12:29
$begingroup$
math.stackexchange.com/questions/437523/…
$endgroup$
– lab bhattacharjee
Jan 10 at 12:29
add a comment |
2 Answers
2
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oldest
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$begingroup$
Evaluate
$$int_0^1frac{1-(1-x)^n}xdx$$
in two different ways.
First way:
$$1-(1-x)^n=1-sum_{k=0}^n(-1)^kbinom nkx^k=sum_{k=1}^n(-1)^{k+1}binom nkx^k$$
so
$$int_0^1frac{1-(1-x)^n}xdx=sum_{k=1}^n(-1)^{k+1}binom nkint_0^1x^{k-1}dx=sum_{k=1}^n(-1)^{k+1}binom nkfrac1k$$
$$=frac{binom n1}1-frac{binom n2}2+frac{binom n3}3-cdots+(-1)^{n+1}frac{binom nn}n.$$
Second way: Substituting $u=1-x$,
$$int_0^1frac{1-(1-x)^n}xdx=int_0^1frac{1-u^n}{1-u}du=int_0^1(1+u+u^2+cdots+u^{n-1})du$$$$=frac11+frac12+frac13+cdots+frac1n.$$
answered Jan 10 at 13:11
bofbof
52.6k559121
$endgroup$
add a comment |
$begingroup$
In other words, you are saying that
begin{equation}
sum_{k=1}^n dfrac{left(-1right)^{k-1}}{k} dbinom{n}{k} = dfrac{1}{1} + dfrac{1}{2} + cdots + dfrac{1}{n}
end{equation}
(because your $C_k$ are precisely the binomial coefficients $dbinom{n}{k}$).
This is a fairly known identity. The one place I remember seeing a proof (because I wrote it) is Exercise 3.19 in my Notes on the combinatorial fundamentals of algebra, version of 10th of January 2019. I suspect it's been on math.stackexchange a few times already.
(If you're looking for a hint: Induction on $n$.)
answered Jan 10 at 12:18
darij grinbergdarij grinberg
11.5k33168
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
Evaluate
$$int_0^1frac{1-(1-x)^n}xdx$$
in two different ways.
First way:
$$1-(1-x)^n=1-sum_{k=0}^n(-1)^kbinom nkx^k=sum_{k=1}^n(-1)^{k+1}binom nkx^k$$
so
$$int_0^1frac{1-(1-x)^n}xdx=sum_{k=1}^n(-1)^{k+1}binom nkint_0^1x^{k-1}dx=sum_{k=1}^n(-1)^{k+1}binom nkfrac1k$$
$$=frac{binom n1}1-frac{binom n2}2+frac{binom n3}3-cdots+(-1)^{n+1}frac{binom nn}n.$$
Second way: Substituting $u=1-x$,
$$int_0^1frac{1-(1-x)^n}xdx=int_0^1frac{1-u^n}{1-u}du=int_0^1(1+u+u^2+cdots+u^{n-1})du$$$$=frac11+frac12+frac13+cdots+frac1n.$$
answered Jan 10 at 13:11
bofbof
52.6k559121
$endgroup$
add a comment |
$begingroup$
Evaluate
$$int_0^1frac{1-(1-x)^n}xdx$$
in two different ways.
First way:
$$1-(1-x)^n=1-sum_{k=0}^n(-1)^kbinom nkx^k=sum_{k=1}^n(-1)^{k+1}binom nkx^k$$
so
$$int_0^1frac{1-(1-x)^n}xdx=sum_{k=1}^n(-1)^{k+1}binom nkint_0^1x^{k-1}dx=sum_{k=1}^n(-1)^{k+1}binom nkfrac1k$$
$$=frac{binom n1}1-frac{binom n2}2+frac{binom n3}3-cdots+(-1)^{n+1}frac{binom nn}n.$$
Second way: Substituting $u=1-x$,
$$int_0^1frac{1-(1-x)^n}xdx=int_0^1frac{1-u^n}{1-u}du=int_0^1(1+u+u^2+cdots+u^{n-1})du$$$$=frac11+frac12+frac13+cdots+frac1n.$$
answered Jan 10 at 13:11
bofbof
52.6k559121
$endgroup$
add a comment |
$begingroup$
Evaluate
$$int_0^1frac{1-(1-x)^n}xdx$$
in two different ways.
First way:
$$1-(1-x)^n=1-sum_{k=0}^n(-1)^kbinom nkx^k=sum_{k=1}^n(-1)^{k+1}binom nkx^k$$
so
$$int_0^1frac{1-(1-x)^n}xdx=sum_{k=1}^n(-1)^{k+1}binom nkint_0^1x^{k-1}dx=sum_{k=1}^n(-1)^{k+1}binom nkfrac1k$$
$$=frac{binom n1}1-frac{binom n2}2+frac{binom n3}3-cdots+(-1)^{n+1}frac{binom nn}n.$$
Second way: Substituting $u=1-x$,
$$int_0^1frac{1-(1-x)^n}xdx=int_0^1frac{1-u^n}{1-u}du=int_0^1(1+u+u^2+cdots+u^{n-1})du$$$$=frac11+frac12+frac13+cdots+frac1n.$$
answered Jan 10 at 13:11
bofbof
52.6k559121
$endgroup$
Evaluate
$$int_0^1frac{1-(1-x)^n}xdx$$
in two different ways.
First way:
$$1-(1-x)^n=1-sum_{k=0}^n(-1)^kbinom nkx^k=sum_{k=1}^n(-1)^{k+1}binom nkx^k$$
so
$$int_0^1frac{1-(1-x)^n}xdx=sum_{k=1}^n(-1)^{k+1}binom nkint_0^1x^{k-1}dx=sum_{k=1}^n(-1)^{k+1}binom nkfrac1k$$
$$=frac{binom n1}1-frac{binom n2}2+frac{binom n3}3-cdots+(-1)^{n+1}frac{binom nn}n.$$
Second way: Substituting $u=1-x$,
$$int_0^1frac{1-(1-x)^n}xdx=int_0^1frac{1-u^n}{1-u}du=int_0^1(1+u+u^2+cdots+u^{n-1})du$$$$=frac11+frac12+frac13+cdots+frac1n.$$
answered Jan 10 at 13:11
bofbof
52.6k559121
edited Jan 10 at 23:28
answered Jan 10 at 13:11
bofbof
52.6k559121
answered Jan 10 at 13:11
bofbof
52.6k559121
answered Jan 10 at 13:11
bofbof
52.6k559121
52.6k559121
add a comment |
add a comment |
$begingroup$
In other words, you are saying that
begin{equation}
sum_{k=1}^n dfrac{left(-1right)^{k-1}}{k} dbinom{n}{k} = dfrac{1}{1} + dfrac{1}{2} + cdots + dfrac{1}{n}
end{equation}
(because your $C_k$ are precisely the binomial coefficients $dbinom{n}{k}$).
This is a fairly known identity. The one place I remember seeing a proof (because I wrote it) is Exercise 3.19 in my Notes on the combinatorial fundamentals of algebra, version of 10th of January 2019. I suspect it's been on math.stackexchange a few times already.
(If you're looking for a hint: Induction on $n$.)
answered Jan 10 at 12:18
darij grinbergdarij grinberg
11.5k33168
$endgroup$
add a comment |
$begingroup$
In other words, you are saying that
begin{equation}
sum_{k=1}^n dfrac{left(-1right)^{k-1}}{k} dbinom{n}{k} = dfrac{1}{1} + dfrac{1}{2} + cdots + dfrac{1}{n}
end{equation}
(because your $C_k$ are precisely the binomial coefficients $dbinom{n}{k}$).
This is a fairly known identity. The one place I remember seeing a proof (because I wrote it) is Exercise 3.19 in my Notes on the combinatorial fundamentals of algebra, version of 10th of January 2019. I suspect it's been on math.stackexchange a few times already.
(If you're looking for a hint: Induction on $n$.)
answered Jan 10 at 12:18
darij grinbergdarij grinberg
11.5k33168
$endgroup$
add a comment |
$begingroup$
In other words, you are saying that
begin{equation}
sum_{k=1}^n dfrac{left(-1right)^{k-1}}{k} dbinom{n}{k} = dfrac{1}{1} + dfrac{1}{2} + cdots + dfrac{1}{n}
end{equation}
(because your $C_k$ are precisely the binomial coefficients $dbinom{n}{k}$).
This is a fairly known identity. The one place I remember seeing a proof (because I wrote it) is Exercise 3.19 in my Notes on the combinatorial fundamentals of algebra, version of 10th of January 2019. I suspect it's been on math.stackexchange a few times already.
(If you're looking for a hint: Induction on $n$.)
answered Jan 10 at 12:18
darij grinbergdarij grinberg
11.5k33168
$endgroup$
In other words, you are saying that
begin{equation}
sum_{k=1}^n dfrac{left(-1right)^{k-1}}{k} dbinom{n}{k} = dfrac{1}{1} + dfrac{1}{2} + cdots + dfrac{1}{n}
end{equation}
(because your $C_k$ are precisely the binomial coefficients $dbinom{n}{k}$).
This is a fairly known identity. The one place I remember seeing a proof (because I wrote it) is Exercise 3.19 in my Notes on the combinatorial fundamentals of algebra, version of 10th of January 2019. I suspect it's been on math.stackexchange a few times already.
(If you're looking for a hint: Induction on $n$.)
answered Jan 10 at 12:18
darij grinbergdarij grinberg
11.5k33168
answered Jan 10 at 12:18
darij grinbergdarij grinberg
11.5k33168
answered Jan 10 at 12:18
darij grinbergdarij grinberg
11.5k33168
answered Jan 10 at 12:18
darij grinbergdarij grinberg
11.5k33168
11.5k33168
add a comment |
add a comment |
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$begingroup$
math.stackexchange.com/questions/437523/…
$endgroup$
– lab bhattacharjee
Jan 10 at 12:29