Find $n\in N$ for which $2*[\frac{1^2}{2}]+2^2*[\frac{2^2}{3}]+...+2^n*[\frac{n^2}{ n+1}]$

Question

Question

Find $n\in N$ for which

$$2 \times \left[\frac{1^2}{2}\right] + 2^2 \times \left[\frac{2^2}{3}\right] + ... + 2^n \times \left[\frac{n^2}{n+1}\right] = 2^{2025} \times 2022 + 4$$

where $[a]$ is integer part function applied on $a$ (the whole part of $a$)

Idea

I tried getting $2^n \times \left[\frac{n^2}{n+1}\right]$ to a more friendly form, but honestly got to nothing useful.

By brute force, I got to the conclusion that $\left[\frac{n^2}{n+1}\right]$ is always increasing by one... again, I tried demonstrating it but I don't know how.

I hope one of you can help me! Thank you!

Answer 1

We want to solve $$2^2\cdot 1+2^3\cdot 2+ 2^4\cdot 3+…+2^{n}(n-1)=2^{2025}\cdot 2022+4.$$

Let us calculate the left hand side. We must add a lot of powers of $2$. Arrange them in a triangle:

$$\begin{array}{ccccc} 2^2 & 2^3 & 2^4 & … & 2^n \\ & 2^3 & 2^4 & … & 2^n \\ & & 2^4 & … & 2^n \\ & & & … & \\ & & & & 2^n \end{array}$$

It is easy to calculate the sum in each row (using $2^k=(2^{k-1}+…+2^3+2^2+2^1)+2$):

$$\begin{array}{ccccc|c} 2^2 & 2^3 & 2^4 & … & 2^n & 2^{n+1}-2^2\\ & 2^3 & 2^4 & … & 2^n & 2^{n+1}-2^3\\ & & 2^4 & … & 2^n & 2^{n+1}-2^4\\ & & & … & &\\ & & & & 2^n & 2^{n+1}-2^n \end{array}$$

Now we summarize the last column: $$2^{n+1}\cdot(n-1)-(2^{n+1}-2^2)= 2^{n+1}\cdot(n-2)+4.$$

It is clear now that $$n=2024.$$

Answer 2

$$\left\lfloor{n^2\over n+1}\right\rfloor=\left\lfloor n-1 +{1\over n+1}\right\rfloor=n-1$$

So we get $$S=2^2\cdot1 +2^3\cdot2+\cdots+2^n(n-1)$$

Now from the GP formula we know $$ 2+2^2+\cdots2^n=2^{n+1}-2 $$

Adding this to S get

$$ S+2^{n+1}-2=2+2^2\cdot2+2^3\cdot3\cdots2^{n-1}(n-1)+2^n\cdot n\\ 2S+2^{n+2}-4=2^2+2^3\cdot2+\cdots2^n(n-1)+2^{n+1}n $$

You can notice rhs looks a lot like S

$$ 2S+2^{n+2}-4=S+2^{n+1}n $$

Rearranging we get $$ S=2^{n+1}(n-2)-4=2^{2025}2022-4 $$

So clearly n=2024

Answer 3

As already pointed out in comments, $\left\lfloor\dfrac{n^2}{n+1}\right\rfloor$ is just $n-1$ in disguise.

And you are asked to calculate $\displaystyle S=\sum\limits_{k=1}^n(k-1)2^k$

Notice that if we change $2$ by $x$ we get

$\displaystyle\begin{align} S(x)&=\sum\limits_{k=1}^n(k-1)x^k=x^2\sum\limits_{k=1}^n(k-1)x^{k-2}=x^2\sum\limits_{k=1}^n\Big(x^{k-1}\Big)'=x^2\Bigg(\sum\limits_{k=1}^n x^{k-1}\Bigg)'\\\\ &=x^2\Bigg(\sum\limits_{k=0}^{n-1} x^k\Bigg)'=x^2\bigg(\frac{x^n-1}{x-1}\bigg)'=\frac{(nx^2-nx-x^2)\,x^n+x^2}{(x-1)^2}\end{align}$

Note: sums are finite so exchanging sum and derivation is not an issue, just regular stuff expressed in condensed form.

Now of course we substitute back $x=2$ to get

$S(2)=(2n-4)\,2^n+4=(n-2)\,2^{n+1}+4$

And $n=2024$ is clearly solution.

Note for the future:

When you will see sums later with terms in $nx^n$ or $n^2x^n$ or $\frac {x^n}n$ and so on, think about derivating once, twice or integrating the geometric $g(x)=\sum\limits_{k=0}^{n-1}x^k=\dfrac{x^n-1}{x-1}$

For instance rewrite

• $nx^n=xnx^{n-1}=x(x^{n})'$

• $n^2x^n=x^2n(n-1)x^{n-2}+xnx^{n-1}=x^2(x^{n})''+x(x^n)'$

• $\dfrac{x^n}{n}=\int x^{n-1}$

And so on...

You will need to pay attention to starting indices though, but apart from that this is the general technique for such sums.