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Question: Find the value of the expression, \[\left( \sum\limits_{x,y,z}{{{\left( x+1 \right)}^{2}}} \right)-{...

Find the value of the expression, (∑x,y,z(x+1)2)−(∑x,y,z(x))2−3\left( \sum\limits_{x,y,z}{{{\left( x+1 \right)}^{2}}} \right)-{{\left( \sum\limits_{x,y,z}{\left(x \right)} \right)}^{2}}-3 is
(a) 2[∑x,y,zx−∑x,y,zxy]2\left[ \sum\limits_{x,y,z}{x}-\sum\limits_{x,y,z}{xy} \right]
(b) 3[∑x,y,zx2−∑x,y,zx]3\left[ \sum\limits_{x,y,z}{{{x}^{2}}}-\sum\limits_{x,y,z}{x} \right]
(c) 2[∑x,y,zxy−∑x,y,zx2]2\left[ \sum\limits_{x,y,z}{xy}-\sum\limits_{x,y,z}{{{x}^{2}}} \right]
(d) 3[∑x,y,zx2−∑x,y,zxy]3\left[ \sum\limits_{x,y,z}{{{x}^{2}}}-\sum\limits_{x,y,z}{xy} \right]

Explanation

Solution

For solving this problem we expand the given summations and evaluate the given question and try to convert it to one of the options. Here, the given summation is not the sum of numbers from x=1x=1 to x=nx=n . The summations ∑x,y,zx\sum\limits_{x,y,z}{x} are called cyclic summations, which works on cycles of (x,y,z)\left( x,y,z \right) . That is ∑x,y,zx=x+y+z\sum\limits_{x,y,z}{x}=x+y+z . This means that in each step we replace ′x′'x' by ′y′'y' and ′y′'y' by ′z′'z' and ′z′'z' by ′x′'x' until the cycle completes and add them.

Complete step-by-step answer:
Let us assume that the given question as
S=(∑x,y,z(x+1)2)−(∑x,y,z(x))2−3S=\left( \sum\limits_{x,y,z}{{{\left( x+1 \right)}^{2}}} \right)-{{\left( \sum\limits_{x,y,z}{\left( x \right)} \right)}^{2}}-3
We know that the summations ∑x,y,zx\sum\limits_{x,y,z}{x} are called cyclic summations, which works on cycles of (x,y,z)\left( x,y,z \right) . That is ∑x,y,zx=x+y+z\sum\limits_{x,y,z}{x}=x+y+z .
This means that in each step we replace ′x′'x' by ′y′'y' and ′y′'y' by ′z′'z' and ′z′'z' by ′x′'x' until the cycle completes and adds them.
Now by replacing ′x′'x' by ′y′'y' and ′y′'y' by ′z′'z' and ′z′'z' by ′x′'x' until the cycle completes we get
⇒S=[(x+1)2+(y+1)2+(z+1)2]−[x+y+z]2−3.....equation(i)\Rightarrow S=\left[ {{\left( x+1 \right)}^{2}}+{{\left( y+1 \right)}^{2}}+{{\left( z+1 \right)}^{2}} \right]-{{\left[ x+y+z \right]}^{2}}-3.....equation(i)
We know that the formula (a+b)2=a2+2ab+b2{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} and
(a+b+c)2=a2+b2+c2+2(ab+bc+ca){{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2\left( ab+bc+ca \right) .
By using these formulas in above equation (i) we will get
⇒S=[x2+y2+z2+2(x+y+z)+3]−[x2+y2+z2+2(xy+yz+zx)]−3\Rightarrow S=\left[ {{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2\left( x+y+z \right)+3 \right]-\left[ {{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2\left( xy+yz+zx \right) \right]-3
By cancelling the common terms in above equation we will get

& \Rightarrow S=2\left( x+y+z \right)-2\left( xy+yz+zx \right) \\\ & \Rightarrow S=2\left[ \left( x+y+z \right)-\left( xy+yz+zx \right) \right] \\\ \end{aligned}$$ Here, we got the cyclic summation of $$x,y,z$$ . So by converting the above equation into the cyclic summation of $$x,y,z$$ we will get $$\Rightarrow S=2\left[ \sum\limits_{x,y,z}{x}-\sum\limits_{x,y,z}{xy} \right]$$ Therefore, we can write the given question as $$\left( \sum\limits_{x,y,z}{{{\left( x+1 \right)}^{2}}} \right)-{{\left( \sum\limits_{x,y,z}{\left( x \right)} \right)}^{2}}-3=2\left[ \sum\limits_{x,y,z}{x}-\sum\limits_{x,y,z}{xy} \right]$$ So, option (a) is the correct answer. **So, the correct answer is “Option A”.** **Note:** Students need to keep in mind that the given summations are cyclic summations not the normal summations. Due to over reading they may consider that $$\sum{x}=\dfrac{x\left( x+1 \right)}{2}$$ which is not correct in this question. So, we need to consider $$\sum\limits_{x,y,z}{x}=x+y+z$$ . This needs to be taken care and should be careful in the calculations part.