If y=1+2x+4x2+8x3+…i) Show that x=y−12yii) Find the interval in whicn the series is convergent

Step 1y=1+2x+4x2+8x3+…i)It is a G.p series(geometric progressian)Sum of series 11−2x[[fe1+a+ar+ar1+....],[S∞=a1r]]]⇒y=11−2x1−2n=1y2n=1−1y2n=y−1yx=y−12y

Step 2ii) Since the series is of the from∑n=0∞2nxnthen1R=lim2n+12n=2&gt;1∴ the series divergeshence rhese is no anterval in which the series is convergent

Step 1y=1+2x+4x2+8x3+…i)It is a G.p series(geometric progressian)Sum of series 11−2x[[fe1+a+ar+ar1+....],[S∞=a1r]]]⇒y=11−2x1−2n=1y2n=1−1y2n=y−1yx=y−12y

Step 2ii) Since the series is of the from∑n=0∞2nxnthen1R=lim2n+12n=2&gt;1∴ the series divergeshence rhese is no anterval in which the series is convergent

Solved: "If y=1+2x+4x2+8x3+…i) Show that x=y−12yii) Find the interval..."

High School
Algebra
Pre-Algebra
Equations, expressions, and inequalitie

Poemslore8ye

Answered question

2022-04-22

y = 1 + 2 x + 4 x_{2} + 8 x_{3} + \dots

i) Show that

x = \frac{y - 1}{2 y}

ii) Find the interval in whicn the series is convergent

Answer & Explanation

Jamison Newman

Beginner2022-04-23Added 10 answers

Step 1

y = 1 + 2 x + 4 x_{2} + 8 x_{3} + \dots

i)
It is a G.p series(geometric progressian)
Sum of series

\frac{1}{1 - 2 x}

[[f e 1 + a + a r + a r^{1} + . . . .], [S_{\infty} = \frac{a}{1 r}]]]

\Rightarrow y = \frac{1}{1 - 2 x}

1 - 2 n = \frac{1}{y}

2 n = 1 - \frac{1}{y}

2 n = \frac{y - 1}{y}

x = \frac{y - 1}{2 y}

aimadorsozf

Beginner2022-04-24Added 10 answers

Step 2
ii) Since the series is of the from

\sum_{n = 0}^{\infty} 2^{n} x^{n}

then

\frac{1}{R} = lim \frac{2^{n + 1}}{2^{n}} = 2 > 1

∴

the series diverges
hence rhese is no anterval in which the series is convergent

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