We read your photo as:

Given the infinite series y = racx2 + racx24 + racx38 + racx416 + racx532 + racx664 + racx7128 + racx8256 + racx9512 + racx101024 + racx112048 + racx124096 + racx138192 + racx1416384 + racx1532768 + r

Mathematics

ScanSolve AI Answer

Answers

2 parts
8

x=2y1+yx = \frac{2y}{1+y}

9

x=3y2(1+y)x = \frac{3y}{2(1+y)}

Step-by-step solution

8 steps

Here are the solutions for both questions.

Part 8· 4 steps

Given the infinite series y=x2+x24+x38+…y = \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{8} + \dots and 0<x<20 < x < 2, we need to prove that x=2y1+yx = \frac{2y}{1+y}.

  1. Identify the first term (aa) and the common ratio (rr) of the geometric series.

    The first term is a=x2a = \frac{x}{2}.
    The common ratio is r=x2/4x/2=x24⋅2x=x2r = \frac{x^2/4}{x/2} = \frac{x^2}{4} \cdot \frac{2}{x} = \frac{x}{2}.
    The condition 0<x<20 < x < 2 ensures that ∣r∣=∣x2∣<1|r| = \left|\frac{x}{2}\right| < 1, so the series converges.

  2. Use the formula for the sum of an infinite geometric series, S∞=a1−rS_\infty = \frac{a}{1-r}.

    y=x21−x2y = \frac{\frac{x}{2}}{1 - \frac{x}{2}}
  3. Simplify the expression for yy.

    y=x22−x2y = \frac{\frac{x}{2}}{\frac{2-x}{2}} y=x2⋅22−xy = \frac{x}{2} \cdot \frac{2}{2-x} y=x2−xy = \frac{x}{2-x}
  4. Rearrange the equation to solve for xx in terms of yy.

    Multiply both sides by (2−x)(2-x):

    y(2−x)=xy(2-x) = x

    Distribute yy:

    2y−xy=x2y - xy = x

    Move all terms containing xx to one side:

    2y=x+xy2y = x + xy

    Factor out xx:

    2y=x(1+y)2y = x(1+y)

    Divide by (1+y)(1+y) to solve for xx:

    x=2y1+yx = \frac{2y}{1+y}

    This proves the required relationship.

    x=2y1+y\boxed{x = \frac{2y}{1+y}}

Part 9

x=3y2(1+y)x = \frac{3y}{2(1+y)}

Given the infinite series y=23x+49x2+827x3+…y = \frac{2}{3}x + \frac{4}{9}x^2 + \frac{8}{27}x^3 + \dots and 0<x<320 < x < \frac{3}{2}, we need to show that x=3y2(1+y)x = \frac{3y}{2(1+y)}.

  1. Identify the first term (aa) and the common ratio (rr) of the geometric series.

    The first term is a=23xa = \frac{2}{3}x.
    The common ratio is r=49x223x=4x29⋅32x=2x3r = \frac{\frac{4}{9}x^2}{\frac{2}{3}x} = \frac{4x^2}{9} \cdot \frac{3}{2x} = \frac{2x}{3}.
    The condition 0<x<320 < x < \frac{3}{2} ensures that ∣r∣=∣2x3∣<1|r| = \left|\frac{2x}{3}\right| < 1, so the series converges.

  2. Use the formula for the sum of an infinite geometric series, S∞=a1−rS_\infty = \frac{a}{1-r}.

    y=23x1−23xy = \frac{\frac{2}{3}x}{1 - \frac{2}{3}x}
  3. Simplify the expression for yy.

    y=2x33−2x3y = \frac{\frac{2x}{3}}{\frac{3-2x}{3}} y=2x3⋅33−2xy = \frac{2x}{3} \cdot \frac{3}{3-2x} y=2x3−2xy = \frac{2x}{3-2x}
  4. Rearrange the equation to solve for xx in terms of yy.

    Multiply both sides by (3−2x)(3-2x):

    y(3−2x)=2xy(3-2x) = 2x

    Distribute yy:

    3y−2xy=2x3y - 2xy = 2x

    Move all terms containing xx to one side:

    3y=2x+2xy3y = 2x + 2xy

    Factor out 2x2x:

    3y=2x(1+y)3y = 2x(1+y)

    Divide by 2(1+y)2(1+y) to solve for xx:

    x=3y2(1+y)x = \frac{3y}{2(1+y)}

    This shows the required relationship.

    x=3y2(1+y)\boxed{x = \frac{3y}{2(1+y)}}
Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp