If the sum of the series $1.2.4+2.3.7+3.4.10+\dots$ to $n$ terms is $\frac{n(n+1)}{12}\left(a{n}^2+bn+c\right)$ then find the value of $a+b+c$

Find the sum of the following series to $n$ terms:

$1+(1+2)+(1+2+3)+(1+2+3+4)+\dots$

Find the sum of the following series to $n$ terms:

$1\times 2+2\times 3+3\times 4+4\times 5+\dots$

If the sum of the series $3\times 1^2+5\times 2^2+7\times 3^2+\dots$ to $n$ terms is $\frac{n(n+1)}{6}\left(a{n}^2+bn+c\right)$ then find $a+b+c$.

 The sum of the series with $n^{th}$ term $2n^3+3n^2-1$ is $\frac{n}{2}\left(n^3+k{n}^2+kn-1\right)$. Hence, find $k$.

The sum of the series whose $n^{th}$ term is $n^3-3^n$ is of the form $\frac{n^2(n+1{)}^2}{a}-\frac{b}{2}\left(3^n-1\right)$, then $a+b=$

If the sum of the series with $n^{th}$ term $n(n+1)(n+4)$ is $\frac{n(n+1)}{12}\left(a{n}^2+bn+c\right)$ then find the value of $a+b+c$.

Find the sum of the series whose $n^{th}$ term is:

$(2n-1{)}^2$

Find the ${20}^{\text{th }}$ term and the sum of $20$ terms of the series:

$2\times 4+4\times 6+6\times 8+\dots$