Given

The first 10 partial sums are as follows:


The rest of the partial sums can be obtained in similar way.
Step-by-step explanation:
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I’ve attached my work below....
I ran out of space so the step 5 the last step is subtract q1 and q3 which step 3 and 4
So 90.9-81.8=9.1
X = 5/6.
The square of 3x+1 is written as (3x+1)². The square of x-2 is (x-2)². Using the information given to us, we want to solve the equation
(3x+1)²=9(x-2)²
(3x+1)(3x+1)=9(x-2)(x-2)
Multiplying the first two binomials, we have:
3x*3x + 1*3x + 1*3x + 1*1 = 9(x-2)(x-2)
9x²+3x+3x+1 = 9(x-2)(x-2)
9x²+6x+1 = 9(x-2)(x-2)
Multiplying the second two binomials, we have:
9x²+6x+1 = 9(x*x-2*x-2*x-2(-2))
9x²+6x+1 = 9(x²-2x-2x+4)
9x²+6x+1 = 9(x²-4x+4)
Using the distributive property gives us
9x²+6x+1 = 9*x²-9*4x+9*4
9x²+6x+1 = 9x²-36x+36
Subtracting 9x² from both sides leaves us
6x+1 = -36x + 36
Adding 36x to both sides we get
42x+1 = 36
Subtracting 1 from both sides we have
42x = 35
Divide both sides by 42:
42x/42 = 35/42
x = 35/42 = 5/6
Solve by completing the square:
x² + 16x + 64 = 13 + 64
(x + 8)² = 77
x + 8 = +/- √77
x = - 8 +/- √77
OR
x = - 8 - √77, - 8 + √77