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Question: Write down and simplify: The \({{\text{8}}^{th}}\)term of \({\left( {1 + 2x} \right)^{ - \dfrac{1}...

Write down and simplify:
The 8th{{\text{8}}^{th}}term of (1+2x)12{\left( {1 + 2x} \right)^{ - \dfrac{1}{2}}}

Explanation

Solution

Hint:General term (i.e.(r+1)th term)\left( {{\text{i}}{\text{.e}}{\text{.}}{{\left( {r + 1} \right)}^{th}}{\text{ term}}} \right)in the expansion of (1+x)n{\left( {1 + x} \right)^n}=Tr+1=n(n1)(n2)............(nr+1)r!xr{{\text{T}}_{r + 1}} = \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)............\left( {n - r + 1} \right)}}{{r!}}{x^r}

Given equation is:
(1+2x)12{\left( {1 + 2x} \right)^{ - \dfrac{1}{2}}}
Now as we know the general term (i.e.(r+1)th term)\left( {{\text{i}}{\text{.e}}{\text{.}}{{\left( {r + 1} \right)}^{th}}{\text{ term}}} \right) in the expansion of (1+x)n{\left( {1 + x} \right)^n} is given as
Tr+1=n(n1)(n2)............(nr+1)r!xr{{\text{T}}_{r + 1}} = \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)............\left( {n - r + 1} \right)}}{{r!}}{x^r}
8th\therefore {8^{th}}Term isT8{{\text{T}}_8}, for r = 7{\text{r = 7}}

T8=[12(121)(122)(123)(124)(125)(126)7!(2x)7] [12(32)(52)(72)(92)(112)(132)7×6×5×4×3×2×1(2x)7] [(1)7.1.3.5.7.9.11.1327.7.6.5.4.3.2.127.x7] [9.11.136.4.2.1](x7)=128748x7=42916x7  \therefore {{\text{T}}_8} = \left[ {\dfrac{{\dfrac{{ - 1}}{2}\left( {\dfrac{{ - 1}}{2} - 1} \right)\left( {\dfrac{{ - 1}}{2} - 2} \right)\left( {\dfrac{{ - 1}}{2} - 3} \right)\left( {\dfrac{{ - 1}}{2} - 4} \right)\left( {\dfrac{{ - 1}}{2} - 5} \right)\left( {\dfrac{{ - 1}}{2} - 6} \right)}}{{7!}}{{\left( {2x} \right)}^7}} \right] \\\ \Rightarrow \left[ {\dfrac{{\dfrac{{ - 1}}{2}\left( {\dfrac{{ - 3}}{2}} \right)\left( {\dfrac{{ - 5}}{2}} \right)\left( {\dfrac{{ - 7}}{2}} \right)\left( {\dfrac{{ - 9}}{2}} \right)\left( {\dfrac{{ - 11}}{2}} \right)\left( {\dfrac{{ - 13}}{2}} \right)}}{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{\left( {2x} \right)}^7}} \right] \\\ \Rightarrow \left[ {\dfrac{{{{\left( { - 1} \right)}^7}.1.3.5.7.9.11.13}}{{{2^7}.7.6.5.4.3.2.1}}{2^7}.{x^7}} \right] \\\ \Rightarrow - \left[ {\dfrac{{9.11.13}}{{6.4.2.1}}} \right]\left( {{x^7}} \right) = - \dfrac{{1287}}{{48}}{x^7} = - \dfrac{{429}}{{16}}{x^7} \\\

So, this is the required 8th{8^{th}}term.

Note: -In such types of questions the key concept is that we have to remember the general term (i.e.(r+1)th term)\left( {{\text{i}}{\text{.e}}{\text{.}}{{\left( {r + 1} \right)}^{th}}{\text{ term}}} \right) which is stated above in the expansion of (1+x)n{\left( {1 + x} \right)^n}, then calculate the required term using this formula and simplify then we will get the required answer.