Question

$\underline{ \tt{question :}} \begin{cases}\bf{ \red{If \: \: the \: seventh \: term \: from \: the \: beginning \: and \: end \: in \: \: the \: binomial \: expansion \: of \: \bigg( \sqrt[3]{2 }{ + \frac{1}{ \sqrt[3]{3} } }^{n} \bigg) \: are equal \: , find \: n}} \end{cases}$
$\underline{ \tt{Answer : }} \begin{cases} \boxed{ \pmb{ \frak{ \blue{n \: = 12}}}} \end{cases}$

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Answer 1

$\: \huge\bf\underline \red{\underline{Answer}}$

$\huge\underline{\overline{\mid{\bold{\blue{\mathcal{n= 12}}\mid}}}}$

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Answer 2

ANSWER:

• The value of n = 12.

GIVEN:

• The seventh term from the beginning and end in the binomial expansion are equal.

TO FIND:

• The number of terms.

EXPLANATION:

$\bigstar\blue{\boxed{\bold{\large{\pink{T_{r+1} = \ ^nC_r\ a^{n-r}b^r}}}}} \\ \\ \\ \dashrightarrow \sf 7^{th}\ term\ from\ the\ beginning \implies r + 1 = 7 \\ \\ \\ \dashrightarrow\sf T_{6+1} = \ ^nC_6\ (\sqrt[3]{2})^{n-6} \bigg( \dfrac{1}{ \sqrt[3]{3} }\bigg) ^6 \\ \\ \\ \dashrightarrow\sf T_{6+1} = \ ^nC_6\ (\sqrt[3]{2})^{n-6}(\sqrt[3]{3} ) ^{ - 6} \\ \\ \\ \sf \dashrightarrow 7^{th}\ term\ from\ the\ end \implies r + 1 = n + 1 - 7 + 1 \\ \\ \\ \sf \implies r + 1 = n - 5 \\ \\ \\ \dashrightarrow \sf T_{n - 6+1} = \ ^nC_{n - 6}\ (\sqrt[3]{2})^{n-n + 6} \bigg( \dfrac{1}{ \sqrt[3]{3} }\bigg) ^{n - 6} \\ \\ \\ \dashrightarrow \sf T_{n - 6+1} = \ ^nC_{n - 6}\ (\sqrt[3]{2})^{6}(\sqrt[3]{3} ) ^{ 6 - n} \\ \\ \\ \dashrightarrow\sf T_{6+1} = T_{n-6+1} \\ \\ \\ \dashrightarrow \sf \dfrac{T_{6+1}}{T_{n-6+1}} = 1 \\ \\ \\ \sf \dashrightarrow \dfrac{\ ^nC_6\ (\sqrt[3]{2})^{n-6}(\sqrt[3]{3} ) ^{ - 6}}{\ ^nC_{n - 6}\ (\sqrt[3]{2})^{6}(\sqrt[3]{3} ) ^{ 6 - n}} = 1$

$\bigstar\red{\boxed{\bold{\large{\green{^nC_r=\dfrac{n!}{r!(n-r)!}}}}}} \\ \\ \\ \dashrightarrow\sf \dfrac{\dfrac{n!}{6!(n-6)!}\ (\sqrt[3]{2})^{n-6}(\sqrt[3]{3} ) ^{ - 6}}{\dfrac{n!}{(n - 6)!(n-n + 6)!}\ (\sqrt[3]{2})^{6}(\sqrt[3]{3} ) ^{ 6 - n}} = 1 \\ \\ \\ \dashrightarrow \sf \dfrac{(\sqrt[3]{2})^{n-6}(\sqrt[3]{3} ) ^{ - 6}}{(\sqrt[3]{2})^{6}(\sqrt[3]{3} ) ^{ 6 - n}} = 1 \\ \\ \\ \dashrightarrow\sf (\sqrt[3]{2})^{n-6 - 6}(\sqrt[3]{3} ) ^{ - 6 - 6 + n} = 1 \\ \\ \\ \dashrightarrow \sf (\sqrt[3]{2})^{n-12}(\sqrt[3]{3} ) ^{n - 12} = 1 \\ \\ \\ \dashrightarrow\sf (\sqrt[3]{6})^{n-12}= 1$

$\dashrightarrow\sf \bigg(6 \bigg)^ {\dfrac{{n-12}}{3} }= {6}^{0}\\ \\ \\ \tt \orange {\star \ Bases \ are \ equal\ so, \ equate \ the \ powers.}\\ \\ \\ \dashrightarrow\sf \dfrac{{n-12}}{3}= 0 \\ \\ \\ \dashrightarrow\sf n-12= 0 \\ \\ \\ \dashrightarrow \sf n = 12$

The value of n = 12.