Amplified Arithmetic

For a fifth degree polynomial $\displaystyle P(x) = \sum_{i=0}^5 {a_ix^i}$ .

Then the coefficient of $x^5$ of its square is given by: $\displaystyle \sum_{i=0}^5 {a_ia_{5-i}}$ . For the given polynomial $a_i = i+1$ .

$\Rightarrow \displaystyle \quad \sum_{i=0}^5 {a_ia_{5-i}} = \sum_{i=1}^6 {i(7-i)} = \sum_{i=1}^6 {(7i-i^2)} = 7\sum_{i=1}^6 {i} - \sum_{i=1}^6 {i^2}$

$= 7 \times \dfrac {(6)(7)}{2} - \dfrac {(6)(7)(13)}{6}= 147-91 = \boxed{56}$

A one-liner solution (using pascal's triangle)

We observe that, the coefficient of $x^5$ in the expansion of $A(x)=(1+2x+3x^2+4x^3+...+6x^5)^2$ and $B(x)=(1+2x+3x^2+4x^3+5x^4+6x^5+7x^6+...+(r+1)x^r+...)^2$ are same, without any contradiction.

Also, we notice that, a gaze at pascal's triangle (we can always form one, whenever required) helps us solve the problem within seconds.

Before we move in front, be sure to check that, the $n$ - $th$ diagonal of our triangle implies the coefficients of respectively $x^0,x^1,x^2,x^3,...$ in the expansion of $\frac1{(1-x)^n}$ . This is an elementary binomial expansion fact. An example:

So, our picture depicts that, $(1-x)^{-3}=1+3x+6x^2+10x^3+15x^4+...$ and $(1-x)^{-4}=1+4x+10x^2+20x^3+35x^4+56x^5+84x^6+...$ . Note that this is just an example. Our solution is precisely short and swift.

Now, we will expand only and only $B(x)$ , to get our answer.

$(1+2x+3x^2+4x^3+5x^4+6x^5+7x^6+...$

$=((1-x)^{-2})^2$

$=(1-x)^{-4}$

$=1+4x+10x^2+20x^3+35x^4+56x^5+84x^6+...$

Thus, we find $\;\boxed{56}$ . $\;$

P.S.: We do not even require to expand anything but prepare a pascal's triangle and find the $6$ - $th$ element of the $5$ - $th$ diagonal. The problem requires a diagram, and then, it turns out to be a one-liner.

Realize that you only get a $x^5$ one time per term so response is $1*6+2*5+3*4+4*3+5*2+6*1=56$ don't exist other way to get a $x^5$