MHB Find closed form expression for a given sum

[anemone]

Find a closed form expression for $$\sum_{k=1}^{n^2}\dfrac{n-\lfloor\sqrt{k-1}\rfloor}{\sqrt{k}+\sqrt{k+1}}$$.

 

[kaliprasad]

Find a closed form expression for $$\sum_{k=1}^{n^2}\dfrac{n-\lfloor\sqrt{k-1}\rfloor}{\sqrt{k}+\sqrt{k+1}}$$.

Spoiler

each term = $(n- \lfloor \sqrt{k-1}\rfloor)(\sqrt{k+1}-\sqrt{k})$

if we take k fron $m^2+1$ to $(m+1)^2$ that is (2m+1) terms we get

$(n- m)(\sqrt{k+1}-\sqrt{k})$

summing from $m^2+1$ to $(m+1)^2$ we get (n-m)



now we need to add (n-m) with m from 1 to n to get the sum as n^2 - n(n+1)/2 = n(n-1)/2

 

[anemone]

Thank you kaliprasad for your clever solution and sorry for the late reply...:(

I will share with you the quite similar but not completely the same solution with you and MHB:

Spoiler

Note that

$\begin{align*}\displaystyle \sum_{k=1}^{n^2}\dfrac{n-\lfloor\sqrt{k-1}\rfloor}{\sqrt{k}+\sqrt{k+1}}&=\sum_{m=1}^{n}\sum_{k=(m-1)^2+1}^{m^2}\dfrac{n-\lfloor\sqrt{k-1}\rfloor}{\sqrt{k}+\sqrt{k+1}}\\&=\sum_{m=1}^{n}\sum_{k=(m-1)^2+1}^{m^2}\dfrac{n-(m-1)}{\sqrt{k}+\sqrt{k+1}}\\&=\sum_{m=1}^{n}(n-m+1)\sum_{k=(m-1)^2+1}^{m^2}\dfrac{1}{\sqrt{k}+\sqrt{k+1}}\\&=\sum_{m=1}^{n}(n-m+1)\sum_{k=(m-1)^2+1}^{m^2}\sqrt{k}-\sqrt{k+1}\\&=\sum_{m=1}^{n}(n-m+1)(m-(m-1))\\&=\sum_{m=1}^{n}(n-m+1)\\&=n(n+1)-\dfrac{n(n+1)}{2}\\&=\dfrac{n(n+1)}{2}\end{align*}$

 

[kaliprasad]

Thank you kaliprasad for your clever solution and sorry for the late reply...:(

I will share with you the quite similar but not completely the same solution with you and MHB:

Spoiler

Note that

$\begin{align*}\displaystyle \sum_{k=1}^{n^2}\dfrac{n-\lfloor\sqrt{k-1}\rfloor}{\sqrt{k}+\sqrt{k+1}}&=\sum_{m=1}^{n}\sum_{k=(m-1)^2+1}^{m^2}\dfrac{n-\lfloor\sqrt{k-1}\rfloor}{\sqrt{k}+\sqrt{k+1}}\\&=\sum_{m=1}^{n}\sum_{k=(m-1)^2+1}^{m^2}\dfrac{n-(m-1)}{\sqrt{k}+\sqrt{k+1}}\\&=\sum_{m=1}^{n}(n-m+1)\sum_{k=(m-1)^2+1}^{m^2}\dfrac{1}{\sqrt{k}+\sqrt{k+1}}\\&=\sum_{m=1}^{n}(n-m+1)\sum_{k=(m-1)^2+1}^{m^2}\sqrt{k}-\sqrt{k+1}\\&=\sum_{m=1}^{n}(n-m+1)(m-(m-1))\\&=\sum_{m=1}^{n}(n-m+1)\\&=n(n+1)-\dfrac{n(n+1)}{2}\\&=\dfrac{n(n+1)}{2}\end{align*}$

Hello Anemone,
Thanks

However In my solution there is a mistake you did not notice

Spoiler

I forgot to add the last value that is n to give n(n+1)/2