BdMO Online Forum

Asif Hossain wrote: ↑

Mon May 10, 2021 10:20 pm

Consider this following proof of a combinatorial identity i.e $n2^{n-1}=\sum_{k=1}^{n}k \binom{n}{k}$
Proof:
$(1+x)^n=\sum_{k=1}^{n} \binom{n}{k} x^k$

Differentiating both sides wrt x gives

$n(1+x)^{n-1}=\sum_{k=1}^{n}k \binom{n}{k} x^{k-1}$

$x:=1 \Rightarrow n2^{n-1}=\sum_{k=1}^{n}k \binom{n}{k}$
How can this be justified for example if i say $x^2 -1=0$ Differentiating both sides wrt x gives $2x=0 \Rightarrow x=0$ which is not true :/.
Sorry if this sound foolish am i missing something?

~Aurn0b~ wrote: ↑

Mon May 10, 2021 11:26 pm

Asif Hossain wrote: ↑

Mon May 10, 2021 10:20 pm

Consider this following proof of a combinatorial identity i.e $n2^{n-1}=\sum_{k=1}^{n}k \binom{n}{k}$
Proof:
$(1+x)^n=\sum_{k=1}^{n} \binom{n}{k} x^k$

Differentiating both sides wrt x gives

$n(1+x)^{n-1}=\sum_{k=1}^{n}k \binom{n}{k} x^{k-1}$

$x:=1 \Rightarrow n2^{n-1}=\sum_{k=1}^{n}k \binom{n}{k}$
How can this be justified for example if i say $x^2 -1=0$ Differentiating both sides wrt x gives $2x=0 \Rightarrow x=0$ which is not true :/.
Sorry if this sound foolish am i missing something?

In the first case both side is function, and true for all real x; where it's not true for $x^2-1=0$