Question

Using sequences and series please give me answer.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:y = \dfrac{1}{1 + x}$

Consider the series

$\rm :\longmapsto\:y + {y}^{2} + {y}^{3} + - - - + {y}^{n}$

Its an Geometric sequence with

• First term, a = y

• Common ratio, r = y

• Number of terms = n

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ Sum of n  terms of an geometric sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{a(1 - {r}^{n})}{1 - r} \: provided \: that \: r \: \ne \: 1}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• Sₙ is the sum of n terms of GP.

• a is the first term of the sequence.

• n is the no. of terms.

• r is the common ratio.

So, using this identity, we get

$\rm :\longmapsto\:S_n = \dfrac{y(1 - {y}^{n} )}{1 - y}$

can be rewritten as

$\rm :\longmapsto\:S_n = \dfrac{1}{1 + x} \times \dfrac{(1 - {y}^{n} )}{1 - \dfrac{1}{1 + x} }$

$\rm :\longmapsto\:S_n = \dfrac{1}{1 + x} \times \dfrac{(1 - {y}^{n} )}{ \dfrac{1 + x - 1}{1 + x} }$

$\rm :\longmapsto\:S_n = \dfrac{1}{1 + x} \times \dfrac{(1 - {y}^{n} )}{ \dfrac{x }{1 + x} }$

$\rm :\longmapsto\:S_n = \dfrac{1 - {y}^{n}}{ x }$

Hence,

$\rm :\longmapsto\:y + {y}^{2} + {y}^{3} + - - + {y}^{n} = \dfrac{1 - {y}^{n}}{ x }$

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Additional Information :-

↝ nᵗʰ term of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• aₙ is the nᵗʰ term.

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common difference.

↝ Sum of n  terms of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• Sₙ is the sum of n terms of AP.

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common difference.