Question

★ Maths Problem !!

1: Find the sum upto $\sf{n^{th}}$ term of the following series :

• $\boxed{\sf{\orange{ 4\: + \: 44 \: + \:444 \: + \:. . . . . . . . . .\:+\: n^{th} term }}}$

• $\boxed{\sf{\orange{ 1 \:+\: \dfrac{1}{2} \:+\: \dfrac{1}{2²} \:+\: \dfrac{1}{2³} \:+\: . . . . . . . . . . \:+\: n^{th} term }}}$

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Answer 1

$\frak{ \maltese \: \: Given :: - } \begin{cases}{ \bullet \: \:{\sf{ \underline{ 4\: + \: 44 \: + \:444 \: + \:. . . . . . . . . .\:+\: upto \: n^{th} term }.}}} \\ \\ \bullet \: \: { \underline{\sf{{ 1 \:+\: \dfrac{1}{2} \:+\: \dfrac{1}{2^2} \:+\: \dfrac{1}{2^3} \:+\: . . . . . . . . . . \:+\: upto \: n^{th} term }.}}} \end{cases}$

✠ Need To Find : We have to find the sum upto $n^{th}\;$term.

$\frak{\maltese\: \:Here : : - } \begin{cases}{ \bullet \: \: \underline{{\textbf{\textsf{S}}} \bf_n \: : \mapsto \: \sf{sum \:of \: { \textbf{ \textsf{ n \: terms. }}}}}} \\{ \bullet \: \: \underline{{\textbf{\textsf{S}}} \bf_{\infty} \: : \mapsto \: \sf{sum \:of \:\infty_{n} { \textbf{ \textsf{\: terms. }}}}}}\\{ \bullet \: \: \underline{{\textbf{\textsf{a}}} \: : \mapsto \: { \textbf{ \textsf{ first \: term. }}}}} \\{ \bullet \: \: \underline{{\textbf{\textsf{r}}} \: : \mapsto \: { \textbf{ \textsf{ common \: ratio. }}}}} \\ { \bullet \: \: \underline{{\textbf{\textsf{G.P.}}} \: : \mapsto \: { \textbf{ \textsf{geometric \: progression. }}}}}. \end{cases}$

${\maltese\: \: \large{ \underline{\underline{ \textsf{ \textbf{Required \: Solution \: (1)}}}}} \: : : -}$

$\frak{ \maltese \: \: Formula \: Used : : - } \begin{cases}\bullet{ \textbf{ \textsf{{\: \: Sum \: of \: n \: terms \: of \: G.P}}}_{ \sf(geometric \: progression)}}. \\\ \bigstar \: \: \underline{\boxed{\bigg \lbrack{ \sf{S_{n}(sum \: of \: n \: terms) \: = \: {\frac{a_{(first \: term)}(r{}^{n}_{(common \: ratio)} - 1) }{(r_{(common \: ratio)} - 1)} \bigg \rbrack ,r_{(common \: ratio)} \neq \: 1}}}}} \: \: \bigstar \end{cases}$

$\bigstar \: \: { \tt{{ 4\: + \: 44 \: + \:444 \: + \:. . . . . . . . . .\:+ \: n^{th} \: term }}}$

➦ Here, we have to convert the given terms into series to get the result by applying the formula :

${ :\longmapsto \tt \:\:S_{n} \: = \: \frac{4}{1} \: or \: 4 \: (1 + 11 + 111 + \cdots\cdots \: + \: n \: term) }$

${:\longmapsto \tt \:\:S_{n} \: = \: \frac{4}{9} \: (9 + 99 + 999 + \cdots\cdots \: + \: n \: term)}$

${ :\longmapsto \tt \:\:S_{n} \: = \: \frac{4}{9} \: \lbrack(10 - 1) +(10 {}^{2} - 1) + (10 {}^{3} - 1) + \cdots\cdots \: + \: n \: term) \rbrack}$

➦ Then, seperating the terms :

${ :\longmapsto \tt \:\:S_{n} \: = \: \frac{4}{9} \: \lbrack(10 +10 {}^{2} + 10 {}^{3} + \cdots\cdots \: + \: 10 {}^{n} ) - (1 + 1 + 1 + \cdots\cdots \: + \: n \: terms ) \rbrack}$

➦ Here, in the terms we can see that ${\bf{a_{(first \:term)\;}}}$= 10 and ${\bf{r_{(common \:ratio)\;}}}$= 10.

➦ Substituting the values according to the formula and given info. :

${\tt{ : \longmapsto \: \: S_n \: = \: \frac{4}{9} \cdot10 \bigg \lbrack \frac{(10 {}^{n} - 1)}{(10 - 1)} \bigg \rbrack -n (1)}}$

$\underline{\boxed{\sf{ : \implies \: \: S_n \: = \: \frac{4}{9} \bigg \lbrack \frac{10}{9}(10^n - 1)-n \bigg \rbrack -n (1)}}} \: \: \bigstar$

.°. Hence, the sum of 4 + 44 + 444 $\cdots\cdots$ upto n terms is $\\ \bf{: \implies \: \: S_n \: = \: \frac{4}{9} \bigg \lbrack \frac{10}{9}(10^n - 1)-n \bigg \rbrack - n}$

${\maltese\: \: \large{ \underline{\underline{ \textsf{ \textbf{Required \: Solution \: (2)}}}}} \: : : -}$

$\frak{ \maltese \: \: Formula \: Used : : - } \begin{cases}\bullet{ \textbf{ \textsf{{\: \: Sum \: of \: n \: terms \: of \: G.P}}}_{ \sf(geometric \: progression)}}. \\\ \bigstar \: \: \underline{\boxed{\bigg \lbrack{ \sf{S_{\infty}(sum \: of \: \infty_{(n)} \: terms) \: = \: {\frac{a_{(first \: term)}}{(1 - r_{(common \: ratio))}} \bigg \rbrack ,|r|_{(common \: ratio)} \:< 1}}}}} \: \:\bigstar \end{cases}$

$\bigstar\;\:{\sf{{ 1 \:+\: \dfrac{1}{2} \:+\: \dfrac{1}{2^2} \:+\: \dfrac{1}{2^3} \:+\: . . . . . . . . . . \:+\: n^{th} term }.}}$

➦ Substituting the values according to the formula and given info. :

$:\longmapsto{ \tt{S_{\infty}\:=\:\dfrac{1}{1-\frac{1}{2}}}}$

$:\longmapsto{ \tt{S_{\infty}\:=\:\dfrac{1}{\frac{1}{2}}}}$

$:\longmapsto{ \tt{S_{\infty}\:=\:1\times\dfrac{2}{1}}}$

${\underline{\boxed{\sf{:\implies S_{\infty}\:=\:2}}}}\;\:\bigstar$

.°. Hence, the sum of ${\sf{{ 1 \:+\: \dfrac{1}{2} \:+\: \dfrac{1}{2^2} \:+\: \dfrac{1}{2^3} \:+\: . . . . . . . . . . \:+\: n^{th} term }.}}$ is $\\ {\bf{:\implies S_{\infty}\:=\:2}}$

Answer 2

Answer:

2 is the answer

Step-by-step explanation:

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