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The summation equation given in the above problem is as follows:

$ \sum\limits_{x=5}^{n+5}{4\left( x-3 \right)}=A{{n}^{2}}+Bn+C $

We are solving the left hand side of the above equation and then compare its result with the R.H.S of the above equation.

$ \sum\limits_{x=5}^{n+5}{4\left( x-3 \right)} $

Substituting x as 5 then x as 6 and so on till x as n + 5 in the L.H.S of the above equation we get,

$ 4\left( 5-3 \right)+4\left( 6-3 \right)+4\left( 7-3 \right)+4\left( 8-3 \right)+.........+4\left( n+5-3 \right) $

Rearranging the above expression we get,

$ \begin{align}

& 4\left( 5+6+7+8+....+n+5 \right)-\left( n+1 \right)4\left( 3 \right) \\

& =4\left( 5+6+7+8+....+n+n+1+n+2+n+3+n+4+n+5 \right)-12\left( n+1 \right) \\

\end{align} $

Now, writing 1, 2, 3, 4 of n + 1, n + 2, n + 3, n + 4 in front of 5 in the above equation we get,

$ 4\left( 1+2+3+4+5+6+....+n+5n \right)-12\left( n+1 \right) $

We can see from the above expression that:

$ 1+2+3+4+....+n $ are sum of first n natural numbers and we know that sum of first n natural numbers are:

$ \dfrac{n\left( n+1 \right)}{2} $

Substituting the above summation in the above summation expression we get,

$ \begin{align}

& 4\left( \dfrac{n\left( n+1 \right)}{2}+5n \right)-12\left( n+1 \right) \\

& =4\left( \dfrac{{{n}^{2}}+n+10n}{2} \right)-12\left( n+1 \right) \\

& =2\left( {{n}^{2}}+11n \right)-12n-12 \\

\end{align} $

Writing $ {{n}^{2}} $ terms, n terms separately we get,

$ \begin{align}

& 2{{n}^{2}}+22n-12n-12 \\

& =2{{n}^{2}}+10n-12 \\

\end{align} $

R.H.S of the given equation is $ A{{n}^{2}}+Bn+C $ . Comparing the above expression with the R.H.S we get the values of A, B and C.

$ \begin{align}

& A=2; \\

& B=10 \\

& C=-12 \\

\end{align} $

Substituting these values of A, B and C in A + B – C we get,

$ \begin{align}

& A+B-C \\

& =2+10-\left( -12 \right) \\

& =12+12 \\

& =24 \\

\end{align} $

Hence, the value of $ A+B-C $ is equal to 24.

$ \dfrac{n\left( n+1 \right)}{2} $

As you can see that the above expression has a quadratic term in n. And as you can see the R.H.S of the given equation is also quadratic in n so there must be some term in L.H.S which is quadratic in n.