Finding b_n - Binomial theorem problem

[Saitama]

Question:
If $\displaystyle \sum_{r=0}^{2n} a_r(x-2)^r=\sum_{r=0}^{2n} b_r(x-3)^r$ and $a_k=1$ for all $k \geq n$, then show that $b_n={}^{2n+1}C_{n+1}$.

Attempt:
I haven't been able to make any useful attempt on this one. I could rewrite it to:

$$\sum_{r=0}^{n-1} a_r(x-2)^r + (x-2)^n\left(\frac{(x-2)^{n+1}-1}{(x-2)-1}\right)=\sum_{r=0}^{2n} b_r(x-3)^r$$

I am clueless about the next step.

Any help is appreciated. Thanks!

 

[I like Serena]

Question:
If $\displaystyle \sum_{r=0}^{2n} a_r(x-2)^r=\sum_{r=0}^{2n} b_r(x-3)^r$ and $a_k=1$ for all $k \geq n$, then show that $b_n={}^{2n+1}C_{n+1}$.

Attempt:
I haven't been able to make any useful attempt on this one. I could rewrite it to:

$$\sum_{r=0}^{n-1} a_r(x-2)^r + (x-2)^n\left(\frac{(x-2)^{n+1}-1}{(x-2)-1}\right)=\sum_{r=0}^{2n} b_r(x-3)^r$$

I am clueless about the next step.

Any help is appreciated. Thanks!

Hi Pranav!

Did you already expand it for $n=2$ and $n=3$?
Just to see if we can find a pattern?

I don't have a solution yet, but I can see that $b_n$ can be fully deduced from the terms starting from n up to 2n.
All terms below n are irrelevant.Let us define $c_t$ such that the sum is equal to \(\displaystyle \sum_{t=0}^{2n} c_t x^t\).
That is, we have that:
$$\sum_{t=0}^{2n} c_t x^t = \sum_{r=0}^{2n} (x-2)^r=\sum_{s=0}^{2n} b_s(x-3)^s$$

With $a_{2n}=1$, we get that highest power $x^{2n}$ has coefficient:
$$c_{2n}=b_{2n}=a_{2n}=1$$
For $n=0$ we would be done, since $b_n = b_0 = \binom{2\cdot 0 + 1}{0+1} = \binom{1}{1} = 1$.Next is $c_{2n-1}$, which yields:
\begin{array}{lclcl}
c_{2n-1} &=& a_{2n-1} + \binom{2n}{1}a_{2n}(-2) &=& b_{2n-1} + \binom{2n}{1}b_{2n}(-3) \\
c_{2n-1} &=& 1 - 4n &=& b_{2n-1} - 6n \\
b_{2n-1} &=& 2n + 1
\end{array}
For $n=1$ we would be done, since $b_n = b_1 = 2\cdot 1 + 1 = \binom{2\cdot 1 + 1}{1+1} = \binom{3}{2} = 3$.Next is $c_{2n-2}$...
Sorry, that's as far as I got at this time. ;)

 

[Saitama]

Hi Pranav!

Did you already expand it for $n=2$ and $n=3$?
Just to see if we can find a pattern?

I don't have a solution yet, but I can see that $b_n$ can be fully deduced from the terms starting from n up to 2n.
All terms below n are irrelevant.Let us define $c_t$ such that the sum is equal to \(\displaystyle \sum_{t=0}^{2n} c_t x^t\).
That is, we have that:
$$\sum_{t=0}^{2n} c_t x^t = \sum_{r=0}^{2n} (x-2)^r=\sum_{s=0}^{2n} b_s(x-3)^s$$

With $a_{2n}=1$, we get that highest power $x^{2n}$ has coefficient:
$$c_{2n}=b_{2n}=a_{2n}=1$$
For $n=0$ we would be done, since $b_n = b_0 = \binom{2\cdot 0 + 1}{0+1} = \binom{1}{1} = 1$.Next is $c_{2n-1}$, which yields:
\begin{array}{lclcl}
c_{2n-1} &=& a_{2n-1} + \binom{2n}{1}a_{2n}(-2) &=& b_{2n-1} + \binom{2n}{1}b_{2n}(-3) \\
c_{2n-1} &=& 1 - 4n &=& b_{2n-1} - 6n \\
b_{2n-1} &=& 2n + 1
\end{array}
For $n=1$ we would be done, since $b_n = b_1 = 2\cdot 1 + 1 = \binom{2\cdot 1 + 1}{1+1} = \binom{3}{2} = 3$.Next is $c_{2n-2}$...
Sorry, that's as far as I got at this time. ;)

Thanks for your input ILS! :)

But is there any proper way of doing it, I mean by some algebraic manipulation? I am thinking of differentiating both the sides wrt r number of times. I will see if I reach somewhere using that.

 

[Saitama]

I have reached the right answer but I cannot perform a summation I come across midway in the solution.

Differentiating wrt r n times as I said before, I get:
$$\sum_{r=0}^{2n} r(r-1)...(r-n+1)a_r(x-2)^{r-n}=\sum_{r=0}^{2n} r(r-1)...(r-n+1)b_r(x-3)^{r-n}$$
$$\Rightarrow \sum_{r=0}^{2n} \frac{r!}{(r-n)!}a_r(x-2)^{r-n}=\sum_{r=0}^{2n} \frac{r!}{(r-n)!}b_r(x-3)^{r-n}$$
Substituting x=3 and using the fact that $a_k=1$ for all $k\geq n$,
$$\sum_{r=n}^{2n} \frac{r!}{(r-n)!}=n! \cdot b_n$$

I did the summation using Wolfram Alpha but how do I solve the summation manually?

 

[I like Serena]

I have reached the right answer but I cannot perform a summation I come across midway in the solution.

Differentiating wrt r n times as I said before, I get:
$$\sum_{r=0}^{2n} r(r-1)...(r-n+1)a_r(x-2)^{r-n}=\sum_{r=0}^{2n} r(r-1)...(r-n+1)b_r(x-3)^{r-n}$$
$$\Rightarrow \sum_{r=0}^{2n} \frac{r!}{(r-n)!}a_r(x-2)^{r-n}=\sum_{r=0}^{2n} \frac{r!}{(r-n)!}b_r(x-3)^{r-n}$$
Substituting x=3 and using the fact that $a_k=1$ for all $k\geq n$,
$$\sum_{r=n}^{2n} \frac{r!}{(r-n)!}=n! \cdot b_n$$

I did the summation using Wolfram Alpha but how do I solve the summation manually?

Nope. I have no clue.
Looks like you're nicely going in the right direction, seeing how everything collapses.

Btw, your summations should start at n instead of 0 after taking the n^th derivative.

Anyway, I can see that the summation can be reduced to a summation of combinations that add up to another combination.
And I can see that it holds true for a number of examples.
But as yet I have no clue how to prove it.

 

[Saitama]

Nope. I have no clue.
Looks like you're nicely going in the right direction, seeing how everything collapses.

Btw, your summations should start at n instead of 0 after taking the n^th derivative.

Anyway, I can see that the summation can be reduced to a summation of combinations that add up to another combination.
And I can see that it holds true for a number of examples.
But as yet I have no clue how to prove it.

Sorry for the late reply.

Is there any alternative method if there is no way to solve the summation?

Thanks!

 

[alyafey22]

Differentiating wrt r n times as I said before, I get:
$$\sum_{r=0}^{2n} r(r-1)...(r-n+1)a_r(x-2)^{r-n}=\sum_{r=0}^{2n} r(r-1)...(r-n+1)b_r(x-3)^{r-n}$$

You have some typos

You are differentiating with respect to $x$ .

The nth derivative with respect to $x$

$$\sum_{r=n}^{2n} r(r-1)...(r-n+1)a_r(x-2)^{r-n}=\sum_{r=n}^{2n} r(r-1)...(r-n+1)b_r(x-3)^{r-n}$$

 

[alyafey22]

\(\displaystyle \sum_{r=n}^{2n} \frac{r!}{(r-n)!}=n! \cdot b_n\)

\(\displaystyle b_n =\sum_{r=n}^{2n} \frac{r!}{(r-n)!\, n! } \)

\(\displaystyle b_n =\sum_{r=n}^{2n} {n \choose r } \)

we need to prove that

\(\displaystyle \sum_{r=n}^{2n} {r \choose n } = {2n+1 \choose n+1}={2n+1 \choose n}\)

By induction on $n$

for $n = 1$ we have

\(\displaystyle \sum_{r=1}^{2} {r\choose 1 } = {3 \choose 1}\)

Next assume

\(\displaystyle \sum_{r=k}^{2k} {r \choose k } = {2k+1 \choose k}\)

and we need to prove that

\(\displaystyle \sum_{r=k+1}^{2k+2} {r \choose k+1 } = {2k+3 \choose k+1}\)

I will leave the rest for you.

 

[juantheron]

Given $\displaystyle \sum_{r=0}^{2n}a_{r}\left(x-2\right)^r = \sum_{r=0}^{2n}b_{r}\left(x-3\right)^r$ and $a_{k} = 1$ forall $k\geq n$

Let $(x-3) = t$ Then $(x-2) = (t+1)$

$\displaystyle \sum_{r=0}^{2n}a_{r}(1+t)^r = \sum_{r=0}^{2n}b_{r}t^r$

$\displaystyle a_{0}+a_{1}(1+t)+a_{2}(1+t)^2+...+a_{n}(1+t)^n+a_{n+1}(1+t)^{n+1}+...+a_{2n}(1+t)^{2n} = b_{0}+b_{1}t+b_{2}t^2+...+b_{n}t^n+b_{n+1}t^{n+1}+...+b_{2n}t^{2n}$

Now camparing Coefficient of $t^n$ on both side and using $a_{k}=1$ forall $k\geq n$

$\displaystyle \binom{n}{n}+\binom{n+1}{n}+\binom{n+2}{n}+...+\binom{2n}{n} = b_{n}$

$\displaystyle \binom{n+1}{n+1}+\binom{n+1}{n}+\binom{n+2}{n}+...+\binom{2n}{n} = b_{n}$

Now using $\displaystyle \binom{n}{r}+\binom{n}{r-1} = \binom{n+1}{r}$ succesively

$\displaystyle \binom{2n+1}{n+1} = b_{n}$

Here $\displaystyle b_{n} = $ Coefficient of $t^n$ on $\bf{R.H.S}$

 

[I like Serena]

Next assume

\(\displaystyle \sum_{r=k}^{2k} {r \choose k } = {2k+1 \choose k}\)

and we need to prove that

\(\displaystyle \sum_{r=k+1}^{2k+2} {r \choose k+1 } = {2k+3 \choose k+1}\)

I will leave the rest for you.

This is where I got stuck...

However, the approach of jacks looks perfect! :)

 

[alyafey22]

This is where I got stuck...

Agreed , I should have tried it before posting , sorry .

 

[Saitama]

Given $\displaystyle \sum_{r=0}^{2n}a_{r}\left(x-2\right)^r = \sum_{r=0}^{2n}b_{r}\left(x-3\right)^r$ and $a_{k} = 1$ forall $k\geq n$

Let $(x-3) = t$ Then $(x-2) = (t+1)$

$\displaystyle \sum_{r=0}^{2n}a_{r}(1+t)^r = \sum_{r=0}^{2n}b_{r}t^r$

$\displaystyle a_{0}+a_{1}(1+t)+a_{2}(1+t)^2+...+a_{n}(1+t)^n+a_{n+1}(1+t)^{n+1}+...+a_{2n}(1+t)^{2n} = b_{0}+b_{1}t+b_{2}t^2+...+b_{n}t^n+b_{n+1}t^{n+1}+...+b_{2n}t^{2n}$

Now camparing Coefficient of $t^n$ on both side and using $a_{k}=1$ forall $k\geq n$

$\displaystyle \binom{n}{n}+\binom{n+1}{n}+\binom{n+2}{n}+...+\binom{2n}{n} = b_{n}$

$\displaystyle \binom{n+1}{n+1}+\binom{n+1}{n}+\binom{n+2}{n}+...+\binom{2n}{n} = b_{n}$

Now using $\displaystyle \binom{n}{r}+\binom{n}{r-1} = \binom{n+1}{r}$ succesively

$\displaystyle \binom{2n+1}{n+1} = b_{n}$

Here $\displaystyle b_{n} = $ Coefficient of $t^n$ on $\bf{R.H.S}$

Thanks a lot jacks! :)

BTW, my friend came up with exactly the same procedure this morning and I was going to post it here as soon as I reach home.

EDIT: Looks like you made it a bit lengthy. Rewrite the summation as:

$$\sum_{r=0}^{n-1} a_r(1+t)^{r}+ (1+t)^n\frac{(1+t)^{n+1}-1}{t}=\sum_{r=0}^{2n}b_rt^r$$
$$\sum_{r=0}^{n-1} a_r(1+t)^{r}+ \frac{(1+t)^{2n+1}-(1+t)^n}{t}=\sum_{r=0}^{2n}b_rt^r$$
Now we need the coefficient of $t^n$ on both the sides.

On the left hand side, we find the coefficient of t^(n+1) from (1+t)^(2n+1) as there is a $t$ in the denominator.

Hence, the coefficient of t^n on LHS is $\binom{2n+1}{n+1}$.