Question

If
${(1 + x)}^{n} = C_0 + C_1x + C_2x + - - + C_n {x}^{n}$

prove that

$(1.2)C_2 + (2.3)C_3 + - - + (n - 1)nC_n = n(n - 1) {2}^{n - 2}$
​

Answer 1

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

Given that

$\sf \: {(1 + x)}^{n} =C_0 + C_1x + C_2 {x}^{2} + C_3 {x}^{3} + - - + C_n {x}^{n}$

On differentiating both sides w. r. t. x, we get

$\sf \:\dfrac{d}{dx}{(1 + x)}^{n} =\dfrac{d}{dx}[C_0 + C_1x + C_2 {x}^{2} + C_3 {x}^{3} + - - + C_n {x}^{n}]$

can be rewritten as

$\sf \:\dfrac{d}{dx}{(1 + x)}^{n} =\dfrac{d}{dx}C_0 +\dfrac{d}{dx} C_1x + \dfrac{d}{dx}C_2 {x}^{2} + \dfrac{d}{dx}C_3 {x}^{3} + - - + \dfrac{d}{dx}C_n {x}^{n}$

We know,

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \dfrac{d}{dx} {x}^{n} \: = \: {nx}^{n \: - \: 1} \: }}}$

So, using this, we get

$\sf n {(1 + x)}^{n - 1} = C_1 + 2C_2x + 3C_3 {x}^{2} + - - + nC_n {x}^{n - 1}$

On differentiating both sides w. r. t. x, we get

$\sf \dfrac{d}{dx}n {(1 + x)}^{n - 1} = \dfrac{d}{dx}[C_1 + 2C_2x + 3C_3 {x}^{2} + - - + nC_n {x}^{n - 1}]$

$\sf n(n - 1) {(1 + x)}^{n - 2} = 2C_2 + (2.3)C_3x + - - + (n - 1)nC_n {x}^{n - 2}$

On substituting x = 1, we get

$\sf n(n - 1) {(1 + 1)}^{n - 2} = 2C_2 + (2.3)C_3 + - - + (n - 1)nC_n {1}^{n - 2}$

$\sf n(n - 1) {2}^{n - 2} = (1.2)C_2 + (2.3)C_3 + - - + (n - 1)nC_n {1}^{n - 2}$

Hence, Proved

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Additional Information :-

$\sf \: If \: {(1 + x)}^{n} =C_0 + C_1x + C_2 {x}^{2} + C_3 {x}^{3} + - - + C_n {x}^{n}$

then

$\boxed{ \tt{ \: C_0 + C_2 + C_4 + - - - = C_1 + C_3 + C_5 + - - - }}$

$\boxed{ \tt{ \: C_0 + C_1 + C_2 + C_3 + - - C_n = {2}^{n}}}$

$\boxed{ \tt{ \: C_0 - C_1 + C_2 - C_3 + - - + {( - 1)}^{n} C_n = 0}}$

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

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