Question

prove the following with complete explanation.

I can't understand from solution.

Please quality answer needed.

​

Answer 1

Given Question :-

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

$\sf \: prove \: that \: C_1 + 2C_2 + 3C_3 + - - + nC_n = n {2}^{n - 1}$

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

Given that

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

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

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

We know,

$\boxed{ \tt{ \: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} \: }} \\ \\ \boxed{ \tt{ \: \dfrac{d}{dx}k = 0 \: }}$

So, using this identity, we get

$\rm :\longmapsto\: {n(1 + x)}^{n - 1} = 0 + C_1 + C_2(2x) + C_3(3 {x}^{2}) + - - + C_n( {nx}^{n - 1})$

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

Now, Substitute x = 1, we get

$\rm :\longmapsto\: {n(1 + 1)}^{n - 1} = C_1 + 2C_2(1) + 3C_3{(1)}^{2}+ - - + nC_n{(1)}^{n - 1}$

$\rm :\longmapsto\: {n2}^{n - 1} = C_1 + 2C_2+ 3C_3+ - - + nC_n$

Hence,

$\rm \implies\:\boxed{ \tt{ \: C_1 + 2C_2+ 3C_3+ - - + nC_n = {n2}^{n - 1} \: }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

More to know :-

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