Question

Solve the given term in to simplest form

$\displaystyle \sf { \sum \limits^{n} _{k = 1} \binom{n}{k} \frac{( - 1) ^{k - 1} }{k} }$
​

Answer 1

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

Given summation is

$\rm :\longmapsto\:\displaystyle \sf { \sum \limits^{n} _{k = 1} \binom{n}{k} \frac{( - 1) ^{k - 1} }{k} }$

can also be rewritten as

$\rm \:  =  \: \displaystyle \sf { \sum \limits^{n} _{k = 1} \: ^nC_{k} \frac{( - 1) ^{k - 1} }{k} }$

can be further open as

$\rm \:  =  \: ^nC_{1} - \dfrac{^nC_{2}}{2} + \dfrac{^nC_{3}}{3} + - - - - + \dfrac{ {( - 1)}^{n - 1} \: ^nC_{n}}{n}$

can also be rewritten as

$\rm \:  =  \: C_{1} - \dfrac{C_{2}}{2} + \dfrac{C_{3}}{3} + - - - - + \dfrac{ {( - 1)}^{n - 1} \: C_{n}}{n}$

So, it means

$\rm \:\displaystyle \sf { \sum \limits^{n} _{k = 1} \binom{n}{k} \frac{( - 1) ^{k - 1} }{k} }  =  \: C_{1} - \dfrac{C_{2}}{2} + \dfrac{C_{3}}{3} + - - - + \dfrac{ {( - 1)}^{n - 1} \: C_{n}}{n}$

Now, we know that

$\rm :\longmapsto\: {(1 - x)}^{n} = C_{0} - C_{1}x + C_{2} {x}^{2} + - - + {( - 1)}^{n - 1} C_{n} {x}^{n}$

$\rm :\longmapsto\: {(1 - x)}^{n} =1 - C_{1}x + C_{2} {x}^{2} + - - + {( - 1)}^{n - 1} C_{n} {x}^{n}$

$\rm :\longmapsto\: {(1 - x)}^{n} - 1 = - C_{1}x + C_{2} {x}^{2} + - - + {( - 1)}^{n - 1} C_{n} {x}^{n}$

On dividing by (-1), we get

$\rm :\longmapsto\: 1 - {(1 - x)}^{n} = C_{1}x - C_{2} {x}^{2} - - - + {( - 1)}^{n - 1} C_{n}$

On dividing both sides by x, we get

$\rm :\longmapsto\: \dfrac{1 - {(1 - x)}^{n}}{x} = C_{1} - C_{2} {x}^{} - - - + {( - 1)}^{n - 2} C_{n} {x}^{n - 1}$

Now on integrating both sides between the limits x = 0 and x = 1, we get

$\rm :\longmapsto\:\displaystyle\int_0^1\dfrac{1 - {(1 - x)}^{n}}{x} = \displaystyle\int_0^1C_{1} - C_{2} {x}^{} - - - + {( - 1)}^{n - 2} C_{n} {x}^{n - 1}$

Now, Consider LHS of above integral

$\rm :\longmapsto\:\displaystyle\int_0^1\dfrac{1 - {(1 - x)}^{n}}{x}$

To integrate such integral, we use method of Substitution

So, Substitute

$\rm :\longmapsto\:1 - x = y$

$\rm :\longmapsto\:- dx = dy$

$\rm :\longmapsto\:dx = - dy$

So, above expression can be rewritten as

$\rm \:  =  \: - \displaystyle\int_{1}^0 \frac{1 - {y}^{n}}{1 - y} dy$

$\rm \:  =  \: \displaystyle\int_{0}^1 \frac{1 - {y}^{n}}{1 - y} dy$

Using Binomial theorem, we have

$\rm \:  =  \: \displaystyle\int_{0}^1 (1 + y + {y}^{2} + - - - + {y}^{n - 1}) \: dy$

$\rm \:  =  \: \bigg[y + \dfrac{ {y}^{2} }{2} + \dfrac{ {y}^{3} }{3} + - - - + \dfrac{ {y}^{n} }{n} \bigg]_{0}^1$

$\rm \:  =  \: 1 + \dfrac{1}{2} + \dfrac{1}{3} + - - - + \dfrac{1}{n}$

Now, Consider RHS of above integral

$\rm :\longmapsto \: \displaystyle\int_0^1C_{1} - C_{2} {x}^{} - - - + {( - 1)}^{n - 1} C_{n} {x}^{n - 1}$

$\rm \:  =  \: \bigg[C_{1}x - C_{2}\dfrac{ {x}^{2} }{2} + C_{3}\dfrac{ {x}^{3} }{3} + - - - + {( - 1)}^{n - 1} C_{n} \dfrac{ {x}^{n} }{n} \bigg]_{0}^1$

$\rm \:  =  \: C_{1} - \dfrac{C_{2}}{2} + \dfrac{C_{3}}{3} + - - - - + \dfrac{ {( - 1)}^{n - 1} \: C_{n}}{n}$

Hence, From above we concluded that

$\rm \: \: C_{1} - \dfrac{C_{2}}{2} + \dfrac{C_{3}}{3} + - - - - + \dfrac{ {( - 1)}^{n - 1} \: C_{n}}{n} \\ \rm \:  =  \: 1 + \dfrac{1}{2} + \dfrac{1}{3} + - - - + \dfrac{1}{n}$

$\displaystyle \sf \red{ \rm\implies \:\sum \limits^{n} _{k = 1} \binom{n}{k} \frac{( - 1) ^{k - 1} }{k} = 1 + \dfrac{1}{2} + \dfrac{1}{3} + - - - + \dfrac{1}{n} }$

$\rule{190pt}{2pt}$

Additional Information :-

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

Answer 2

Answer:

$\large\underline{\sf{Solution-}}Solution−</p><p></p><p>Given summation is \\ </p><p></p><p>\rm :\longmapsto\:\displaystyle \sf { \sum \limits^{n} _{k = 1} \binom{n}{k} \frac{( - 1) ^{k - 1} }{k} }:⟼k=1∑n(kn)k(−1)k−1 \\ </p><p></p><p>can also be rewritten as \\ </p><p></p><p>\rm \:  =  \: \displaystyle \sf { \sum \limits^{n} _{k = 1} \: ^nC_{k} \frac{( - 1) ^{k - 1} }{k} } = k=1∑nnCkk(−1)k−1 \\ </p><p></p><p>can be further open as</p><p> \\ </p><p>\rm \:  =  \: ^nC_{1} - \dfrac{^nC_{2}}{2} + \dfrac{^nC_{3}}{3} + - - - - + \dfrac{ {( - 1)}^{n - 1} \: ^nC_{n}}{n} = nC1−2nC2+3nC3+−−−−+n(−1)n−1nCn</p><p></p><p>can also be rewritten as</p><p></p><p>\rm \:  =  \: C_{1} - \dfrac{C_{2}}{2} + \dfrac{C_{3}}{3} + - - - - + \dfrac{ {( - 1)}^{n - 1} \: C_{n}}{n} = C1−2C2+3C3+−−−−+n(−1)n−1Cn</p><p></p><p>So, it means</p><p></p><p>\rm \:\displaystyle \sf { \sum \limits^{n} _{k = 1} \binom{n}{k} \frac{( - 1) ^{k - 1} }{k} }  =  \: C_{1} - \dfrac{C_{2}}{2} + \dfrac{C_{3}}{3} + - - - + \dfrac{ {( - 1)}^{n - 1} \: C_{n}}{n}k=1∑n(kn)k(−1)k−1 = C1−2C2+3C3+−−−+n(−1)n−1Cn</p><p></p><p>Now, we know that</p><p></p><p>\rm :\longmapsto\: {(1 - x)}^{n} = C_{0} - C_{1}x + C_{2} {x}^{2} + - - + {( - 1)}^{n - 1} C_{n} {x}^{n}:⟼(1−x)n=C0−C1x+C2x2+−−+(−1)n−1Cnxn</p><p></p><p>\rm :\longmapsto\: {(1 - x)}^{n} =1 - C_{1}x + C_{2} {x}^{2} + - - + {( - 1)}^{n - 1} C_{n} {x}^{n}:⟼(1−x)n=1−C1x+C2x2+−−+(−1)n−1Cnxn</p><p></p><p>\rm :\longmapsto\: {(1 - x)}^{n} - 1 = - C_{1}x + C_{2} {x}^{2} + - - + {( - 1)}^{n - 1} C_{n} {x}^{n}:⟼(1−x)n−1=−C1x+C2x2+−−+(−1)n−1Cnxn</p><p></p><p>On dividing by (-1), we get</p><p></p><p>\rm :\longmapsto\: 1 - {(1 - x)}^{n} = C_{1}x - C_{2} {x}^{2} - - - + {( - 1)}^{n - 1} C_{n}:⟼1−(1−x)n=C1x−C2x2−−−+(−1)n−1Cn</p><p></p><p>On dividing both sides by x, we get</p><p></p><p>\rm :\longmapsto\: \dfrac{1 - {(1 - x)}^{n}}{x} = C_{1} - C_{2} {x}^{} - - - + {( - 1)}^{n - 2} C_{n} {x}^{n - 1}:⟼x1−(1−x)n=C1−C2x−−−+(−1)n−2Cnxn−1</p><p></p><p>Now on integrating both sides between the limits x = 0 and x = 1, we get</p><p></p><p>\rm :\longmapsto\:\displaystyle\int_0^1\dfrac{1 - {(1 - x)}^{n}}{x} = \displaystyle\int_0^1C_{1} - C_{2} {x}^{} - - - + {( - 1)}^{n - 2} C_{n} {x}^{n - 1}:⟼∫01x1−(1−x)n=∫01C1−C2x−−−+(−1)n−2Cnxn−1</p><p></p><p>Now, Consider LHS of above integral</p><p></p><p>\rm :\longmapsto\:\displaystyle\int_0^1\dfrac{1 - {(1 - x)}^{n}}{x}:⟼∫01x1−(1−x)n</p><p></p><p>To integrate such integral, we use method of Substitution</p><p></p><p>So, Substitute</p><p></p><p>\rm :\longmapsto\:1 - x = y:⟼1−x=y</p><p></p><p>\rm :\longmapsto\:- dx = dy:⟼−dx=dy</p><p></p><p>\rm :\longmapsto\:dx = - dy:⟼dx=−dy</p><p></p><p>So, above expression can be rewritten as</p><p></p><p>\rm \:  =  \: - \displaystyle\int_{1}^0 \frac{1 - {y}^{n}}{1 - y} dy = −∫101−y1−yndy</p><p></p><p>\rm \:  =  \: \displaystyle\int_{0}^1 \frac{1 - {y}^{n}}{1 - y} dy = ∫011−y1−yndy</p><p></p><p>Using Binomial theorem, we have</p><p></p><p>\rm \:  =  \: \displaystyle\int_{0}^1 (1 + y + {y}^{2} + - - - + {y}^{n - 1}) \: dy = ∫01(1+y+y2+−−−+yn−1)dy</p><p></p><p>\rm \:  =  \: \bigg[y + \dfrac{ {y}^{2} }{2} + \dfrac{ {y}^{3} }{3} + - - - + \dfrac{ {y}^{n} }{n} \bigg]_{0}^1 = [y+2y2+3y3+−−−+nyn]01</p><p></p><p>\rm \:  =  \: 1 + \dfrac{1}{2} + \dfrac{1}{3} + - - - + \dfrac{1}{n} = 1+21+31+−−−+n1</p><p></p><p>Now, Consider RHS of above integral</p><p></p><p>\rm :\longmapsto \: \displaystyle\int_0^1C_{1} - C_{2} {x}^{} - - - + {( - 1)}^{n - 1} C_{n} {x}^{n - 1}:⟼∫01C1−C2x−−−+(−1)n−1Cnxn−1</p><p></p><p>\rm \:  =  \: \bigg[C_{1}x - C_{2}\dfrac{ {x}^{2} }{2} + C_{3}\dfrac{ {x}^{3} }{3} + - - - + {( - 1)}^{n - 1} C_{n} \dfrac{ {x}^{n} }{n} \bigg]_{0}^1 = [C1x−C22x2+C33x3+−−−+(−1)n−1Cnnxn]01</p><p></p><p>\rm \:  =  \: C_{1} - \dfrac{C_{2}}{2} + \dfrac{C_{3}}{3} + - - - - + \dfrac{ {( - 1)}^{n - 1} \: C_{n}}{n} = C1−2C2+3C3+−−−−+n(−1)n−1Cn</p><p></p><p>Hence, From above we concluded that</p><p></p><p>\begin{gathered}\rm \: \: C_{1} - \dfrac{C_{2}}{2} + \dfrac{C_{3}}{3} + - - - - + \dfrac{ {( - 1)}^{n - 1} \: C_{n}}{n} \\ \rm \:  =  \: 1 + \dfrac{1}{2} + \dfrac{1}{3} + - - - + \dfrac{1}{n} \end{gathered}C1−2C2+3C3+−−−−+n(−1)n−1Cn = 1+21+31+−−−+n1</p><p></p><p>\displaystyle \sf \red{ \rm\implies \:\sum \limits^{n} _{k = 1} \binom{n}{k} \frac{( - 1) ^{k - 1} }{k} = 1 + \dfrac{1}{2} + \dfrac{1}{3} + - - - + \dfrac{1}{n} }⟹k=1∑n(kn)k(−1)k−1=1+21+31+−−−+n1</p><p></p><p>\rule{190pt}{2pt}</p><p></p><p>Additional Information :-</p><p></p><p>\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}\end{gathered}f(x)ksinxcosxsec2xcosec2xsecxtanxcosecxcotxtanxx1ex∫f(x)dxkx+c−cosx+csinx+ctanx+c−cotx+csecx+c−cosecx+clogsecx+clogx+cex+c</p><p></p><p>$