Question

Find the general term (r+1) term in the expansion of $(4+5x)^{-3/2}$​

Answer 1

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$\sf{(r+1)^{th} \: term \: of \: (4+5x)^{\frac{-3}{2}}}$

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$\sf{(4+5x)^{\frac{-3}{2}}} \implies \: 4 \: . \: \bigg(1 + \dfrac{5x}{4} \bigg)^{ \frac{ - 3}{2} }$

$\sf{(r+1)^{th} \: term \: of \: \: \bigg(1 + \dfrac{5x}{4} \bigg)^{ \frac{ - 3}{2} } }$

$\sf{General mTerm \: for (1+x)^m \implies \boxed{ \boxed{ \sf{T_{r + 1} = \dfrac{m(m-1)(m-2) .... [m - (r-1)]}{r!} \times x^r}}}}$

$\sf m = \dfrac{ - 3}{2} , \: x = \dfrac{5x}{4}$

$\sf{T_{r + 1} = \dfrac{ \dfrac{ - 3}{2} \bigg( \dfrac{ - 3}{2} -1 \bigg) \bigg( \dfrac{ - 3}{2} -2 \bigg) .... \bigg[ \dfrac{ - 3}{2} - r + 1 \bigg]}{r!}} \times { \bigg(\dfrac{ 5x}{4} \bigg) }^{r}$

$\sf{T_{r + 1} = \dfrac{ \dfrac{ - 3}{2} \bigg( \dfrac{ - 5}{2} \bigg) \bigg( \dfrac{ - 7}{2} \bigg) .... \bigg[ \dfrac{ - 1 - 2r}{2} \bigg]}{r!}} \times { \bigg(\dfrac{ 5x}{4} \bigg) }^{r}$

$\sf{T_{r + 1} = {( - 1)}^{r} \times \dfrac{ \dfrac{ 3}{2} \bigg( \dfrac{ 5}{2} \bigg) \bigg( \dfrac{ 7}{2} \bigg) .... \bigg[ \dfrac{ 1 + 2r}{2} \bigg]}{r!}} \times { \bigg(\dfrac{ 5x}{4} \bigg) }^{r}$

$\sf{ T_{r+1} = {( - 1)}^{r} \times \dfrac{3.5.7....(1 + 2r)}{ {2}^{r} \: . \: r !} \times \bigg( \dfrac{5x}{4} \bigg)^{r} }$

$\sf{ T_{r+1} = {( - 1)}^{r} \times \dfrac{3.5.7....(1 + 2r)}{ \: \: r !} \times \bigg( \dfrac{5x}{8} \bigg)^{r} \times {4}^{ \frac{ - 3}{2} } }$

$\sf{(r+1)^{th} \: term \: of \: \: {4}^{ \frac{ - 3}{2} } \: . \: \bigg(1 + \dfrac{5x}{4} \bigg)^{ \frac{ - 3}{2} } } =$

$\sf{ T_{r+1} = {( - 1)}^{r} \times \dfrac{3.5.7....(1 + 2r)}{ \: \: r !} \times \bigg( \dfrac{5x}{8} \bigg)^{r} \times {4}^{ \frac{ - 3}{2} } }$

$\sf{ T_{r+1} = {( - 1)}^{r} \times \dfrac{3.5.7....(1 + 2r)}{ \: \: r !} \times \bigg( \dfrac{5x}{8} \bigg)^{r} \times {( {2}^{2}) }^{ \frac{ - 3}{2} } }$

$\sf{ T_{r+1} = {( - 1)}^{r} \times \dfrac{3.5.7....(1 + 2r)}{ \: \: r !} \times \bigg( \dfrac{5x}{8} \bigg)^{r} \times \dfrac{1}{8} }$

$\boxed{ \boxed{ \sf{ T_{r+1} = {( - 1)}^{r} \times \dfrac{3.5.7....(1 + 2r)}{ \: \: r !} \times \dfrac{(5x)^{r} }{8^{r + 1} } }}}$