Question

# We have given a summation value in which if $\sum\limits_{x=5}^{n+5}{4\left( x-3 \right)}=A{{n}^{2}}+Bn+C$ then $A+B-C$ is equal to:

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Hint: To get the value of the summation, substitute x as 5, then x as 6 and so on till x as n + 5. Then as you can see that in the R.H.S of the equation we are getting ${{n}^{2}}$ which can be found in L.H.S when there are sum of first n natural numbers because sum of first n natural numbers is equal to $\dfrac{n\left( n+1 \right)}{2}$ so arrange the left hand side of the above equation in such a way that we will get the sum of first n natural numbers and then simplify left hand side of the equation. After that compare L.H.S and R.H.S of the equation and find the value of A, B and C and then substitute these values in $A+B-C$ .

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.

Note: You might have thought how we know that we should arrange L.H.S of the summation equation in such a way that the summation is in the form of sum of first n natural numbers because sum of first n natural numbers is equal to:
$\dfrac{n\left( n+1 \right)}{2}$
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.