Amos J. answered • 04/15/17
Tutor
4.9
(34)
Math and Physics
Hi Romano,
I think S = 334.
We can rewrite the three given equations like so:
Σ(n2xn) = x1 + 4x2 + 9x3 + 16x4 + 25x5 + 36x6 + 49x7 = 1
Σ[(n+1)2xn] = 4x1 + 9x2 + 16x3 + 25x4 + 36x5 + 49x6 + 64x7 = 12
Σ[(n+2)2xn] = 9x1 + 16x2 + 25x3 + 36x4 + 49x5 + 64x6 + 81x7 = 123
Where each of those summations go from n=1 to n=7. Let's get get rid of those messy seven-term polynomials and just work with the summations (rewriting below):
Σ(n2xn) = 1
Σ[(n+1)2xn] = 12
Σ[(n+2)2xn] = 123
Let's carry out the square of the (n+1) and (n+2) terms in parentheses:
Σ[(n2 + 2n + 1)xn] = 12
Σ[(n2 + 4n + 4)xn] = 123
Distribute the xn into the parentheses:
Σ[n2xn + 2nxn + xn] = 12
Σ[n2xn + 4nxn + 4xn] = 123
And now we can distribute the summation signs, bringing constant coefficients outside the summations:
Σ(n2xn) + 2Σ(nxn) + Σ(xn) = 12
Σ(n2xn) + 4Σ(nxn) + 4Σ(xn) = 123
But we know from the first given equation that Σ(n2xn) = 1, so the two expressions reduce to:
2Σ(nxn) + Σ(xn) = 11
4Σ(nxn) + 4Σ(xn) = 122
We can expand the first expression by separating 2Σ(nxn) into Σ(nxn) + Σ(nxn), and divide the second expression by 4:
Σ(nxn) + Σ(nxn) + Σ(xn) = 11
Σ(nxn) + Σ(xn) = 30.5
Now we can reduce the first expression, using what we know from the second expression:
Σ(nxn) + 30.5 = 11
Do the subtraction:
Σ(nxn) = -19.5
Now we can go back and figure out that:
-19.5 + Σ(xn) = 30.5
Do the addition:
Σ(xn) = 50
So now we know three things:
Σ(n2xn) = 1
Σ(nxn) = -19.5
Σ(xn) = 50
We can finally figure out how to find S = 16x1 + 25x2 + 36x3 + 49x4 + 64x5 + 81x6 + 100x7
S = Σ[(n+3)2xn]
= Σ[(n2 + 6n + 9)xn]
= Σ(n2xn) + 6Σ(nxn) + 9Σ(xn)
= 1 + 6(-19.5) + 9(50)
S = 334
Hope this has helped!
