Sum to n terms of the following series: $2 \cdot 2 + 6 \cdot 4 + 12 \cdot 8 + 20 \cdot 16 + \cdots $

Question

I able to get the general term or $\large n$th term of the series is: $$U_n=(n^2+n)2^n$$ Now i want to get $S_n$ by the method of difference.That's why I need to make $$U_n=V_r-V_{r-1}$$ My book is providing that $U_n=(2n^2-2n+4)2^n-[2(n-1)^2-2(n-1)+4)2^{n-1}]$.Now my question is how they get $U_n$ like this.Thanks in advance.

Answer 1

We may expect the form of $V_n$ to be $(an^2+bn+c)2^n$, since the polynomial factor of $U_n $ is in degree $2$.

Now

\begin{align} V_n-V_{n-1}&=(an^2+bn+c)2^n-[a(n-1)^2+b(n-1)+c]2^{n-1}\\ &=2^{n-1}[2an^2+2bn+2c-a(n^2-2n+1)-b(n-1)-c]\\ &=2^{n-1}[an^2+(2b+2a-b)n+(2c-a+b-c)]\\ &=2^{n}[\frac a2 n^2+(a+\frac b2)n+\frac{c+b-a}2] \end{align}

Comparing the coefficients, we get $$a=2,b=-2,c=4$$

So $$V_n=(2n^2-2n+4)2^n$$

Answer 2

Just rearrange to obtain \begin{align} n^2+n &= 2n^2-n^2 -2n+3n+4-4\\ &= (2n^2-2n+4)-(n^2-3n+4)\\ &= (2n^2-2n+4) - (n^2-2n+1-n+1+2)\\ &= (2n^2-2n+4) - ((n-1)^2-(n-1)+2)\\ &= (2n^2-2n+4) - \frac{2(n-1)^2-2(n-1)+4}{2}. \end{align} Do you see it now?