$Q.$ Find a NICE form of : $$\sum_{m=1}^{p}\lfloor\log_2m\rfloor$$
APPROACH : We have , $$\lfloor\log_21\rfloor⠀\color{red}{\lfloor\log_22\rfloor}⠀\lfloor\log_23\rfloor⠀\color{red}{\lfloor\log_24\rfloor}⠀\lfloor\log_25\rfloor⠀\lfloor\log_26\rfloor⠀\lfloor\log_27\rfloor⠀\color{red}{\lfloor\log_28\rfloor}⠀........⠀\lfloor\log_2n\rfloor$$ $$0⠀⠀⠀⠀⠀⠀\color{red}1⠀⠀⠀⠀⠀⠀1⠀⠀⠀⠀⠀⠀\color{red}2⠀⠀⠀⠀⠀⠀2⠀⠀⠀⠀⠀2⠀⠀⠀⠀⠀⠀2⠀⠀⠀⠀⠀⠀\color{red}3⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$$
We know that : $$\lfloor\log_22^k\rfloor=k$$
STEP 1 : Sum of terms b/w $\lfloor\log_22^{k-1}\rfloor\to\lfloor\log_22^k\rfloor$ . Number of terms b/w $2^{k-1}\to2^k$ is $2^{k-1}$ .
Hence the sum upto $\underline{\lfloor\log_22^{k-1}\rfloor\to\lfloor\log_22^k\rfloor}$ will be : $$S_{2^{k-1}\to2^k}=\lfloor\log_22^{k-1}\rfloor2^{k-1}+\lfloor\log_22^{k}\rfloor$$ $$S_{2^{k-1}\to2^k}=(k-1)2^{k-1}+k$$
Now sum upto $\underline{\lfloor\log_21\rfloor\to\lfloor\log_22^k\rfloor}$ : $$S_{ \lfloor\log_22^k\rfloor}=k+\underbrace{\sum_{r=1}^{k}(k-1)2^{k-1}}_\phi$$
$\Rightarrow$ Now we'll solve for $\phi$ , $$\phi=0.2^0+1.2^1+2.2^2+3.2^3+.......+(k-1)2^{k-1}$$ $$\phi=2+2.2^2+3.2^3+.......+(k-1)2^{k-1}⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀(1)$$ $$\frac{\phi}{2}=1+2.2+3.2^2+.......+(k-1)2^{k-2}⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀(2)$$
Substracting $eq^n(1)$ from $eq^n(2)$ $$-\frac{\phi}{2}=\underbrace{1+2+2^2+2^3+.......+2^{k-2}}_{n-1⠀terms⠀(GP)}-(k-1)2^{k-1}$$ $$-\frac{\phi}{2}=(2^{k-1}-1)-(k-1)2^{k-1}$$ $$\phi=2((k-1)2^{k-1}-2^{k-1}+1)$$ $$\phi=2^k(k-2)+2$$
$\Rightarrow$ Using the value of $\phi$ in $S_{ \lfloor\log_22^k\rfloor}$ : $$S_{ \lfloor\log_22^k\rfloor}=k+2^k(k-2)+2$$
STEP 2 : Let there be a term $\lfloor\log_2p\rfloor$ b/w $\lfloor\log_22^{k-1}\rfloor\to\lfloor\log_22^k\rfloor$ : $$S_{\lfloor\log_2p\rfloor}=S_{ \lfloor\log_22^k\rfloor}-\lfloor\log_22^k\rfloor-\delta$$
Here $\delta$ is the sum of terms b/w $\lfloor\log_2p\rfloor$ and $\lfloor\log_22^k\rfloor$ . Number of terms b/w $\lfloor\log_2p\rfloor$ and $\lfloor\log_22^k\rfloor$ is $(2^k-p)-1$
Hence $\delta$ : $$\delta=(2^k-p-1)\lfloor\log_2p\rfloor$$
Now , $$S_{\lfloor\log_2p\rfloor}=k+2^k(k-2)+2-\lfloor\log_22^k\rfloor-(2^k-p-1)\lfloor\log_2p\rfloor$$
Here are some important conditions : $$ \lfloor\log_22^k\rfloor=k$$ $$\lfloor\log_22^{k-1}\rfloor=\lfloor\log_2p\rfloor$$ $$\lfloor\log_22^k\rfloor=\lfloor\log_22^{k-1}\rfloor+1=\lfloor\log_2p\rfloor+1$$
FINALLY : after further simplifying , I got :
$$\sum_{m=1}^{p}\lfloor\log_2m\rfloor=\lfloor\log_2p\rfloor(p+1)-2^{\lfloor\log_2p\rfloor+1}+2$$
DOUBT : Is there a simpler OR better approach than this ?
I think this is the same calculation as yours, but much more compact; you might find the technique in the first two lines valuable. \begin{align*} \sum_{m=1}^{p}\lfloor\log_2m\rfloor &= \sum_{m=1}^{p} \sum_{1\le k\le (\log_2m)} 1 \\ &= \sum_{1\le k\le (\log_2 p)} \sum_{m=2^k}^{p} 1 \\ &= \sum_{1\le k\le (\log_2 p)} (p+1-2^k) \\ &= \lfloor\log_2p\rfloor (p+1) - \sum_{k=1}^{\lfloor\log_2p\rfloor}2^k = \lfloor\log_2p\rfloor (p+1) - \big( 2^{\lfloor\log_2p\rfloor+1}-2 \big). \end{align*}