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Question: Prove the following equation: \({C_0} - 2{C_1} + 3{C_2} - 4{C_3} + ....... + {\left( { - 1} \right...

Prove the following equation:
C0−2C1+3C2−4C3+.......+(−1)n(n+1)Cn=0{C_0} - 2{C_1} + 3{C_2} - 4{C_3} + ....... + {\left( { - 1} \right)^n}\left( {n + 1} \right){C_n} = 0

Explanation

Solution

In order to prove the given use the concept of binomial expansion of series. Start with the general formula then make some substitution in the formula according to the problem statement in order to make the series similar to the given one.

Complete step-by-step answer:
We have to prove C0−2C1+3C2−4C3+.......+(−1)n(n+1)Cn=0{C_0} - 2{C_1} + 3{C_2} - 4{C_3} + ....... + {\left( { - 1} \right)^n}\left( {n + 1} \right){C_n} = 0
As we know the general formula for the binomial expansion of the series is given by:
(1+x)n=C0x0+C1x1+C2x2+C3x3+.......+Cnxn{\left( {1 + x} \right)^n} = {C_0}{x^0} + {C_1}{x^1} + {C_2}{x^2} + {C_3}{x^3} + ....... + {C_n}{x^n}
Multiplying both sides of the above equation by x we get:

⇒x×(1+x)n=x×(C0x0+C1x1+C2x2+C3x3+.......+Cnxn) ⇒(1+x)n⋅x=C0x1+C1x2+C2x3+C3x4+.......+Cnxn+1  \Rightarrow x \times {\left( {1 + x} \right)^n} = x \times \left( {{C_0}{x^0} + {C_1}{x^1} + {C_2}{x^2} + {C_3}{x^3} + ....... + {C_n}{x^n}} \right) \\\ \Rightarrow {\left( {1 + x} \right)^n} \cdot x = {C_0}{x^1} + {C_1}{x^2} + {C_2}{x^3} + {C_3}{x^4} + ....... + {C_n}{x^{n + 1}} \\\

Now in order bring the equation something in form of the result let us differentiate the whole equation with respect to x
⇒ddx[(1+x)n⋅x]=ddx[C0x1+C1x2+C2x3+C3x4+.......+Cnxn+1]\Rightarrow \dfrac{d}{{dx}}\left[ {{{\left( {1 + x} \right)}^n} \cdot x} \right] = \dfrac{d}{{dx}}\left[ {{C_0}{x^1} + {C_1}{x^2} + {C_2}{x^3} + {C_3}{x^4} + ....... + {C_n}{x^{n + 1}}} \right]
Now let us open the brackets in order to differentiate the whole term
⇒ddx[(1+x)n⋅x]=ddx[C0x1]+ddx[C1x2]+ddx[C2x3]+ddx[C3x4]+.......+ddx[Cnxn+1]\Rightarrow \dfrac{d}{{dx}}\left[ {{{\left( {1 + x} \right)}^n} \cdot x} \right] = \dfrac{d}{{dx}}\left[ {{C_0}{x^1}} \right] + \dfrac{d}{{dx}}\left[ {{C_1}{x^2}} \right] + \dfrac{d}{{dx}}\left[ {{C_2}{x^3}} \right] + \dfrac{d}{{dx}}\left[ {{C_3}{x^4}} \right] + ....... + \dfrac{d}{{dx}}\left[ {{C_n}{x^{n + 1}}} \right]
As we know the formulas for differentiation are:
ddx(xk)=kxk−1 and ddx(p.q)=pddx(q)+qddx(p)\dfrac{d}{{dx}}\left( {{x^k}} \right) = k{x^{k - 1}}{\text{ and }}\dfrac{d}{{dx}}\left( {p.q} \right) = p\dfrac{d}{{dx}}\left( q \right) + q\dfrac{d}{{dx}}\left( p \right)
Using the above formulas we differentiate the terms in the equation:

⇒(1+x)nddx(x)+xddx[(1+x)n]=C0ddx[x1]+C1ddx[x2]+..... .....C2ddx[x3]+C3ddx[x4]+.......+Cnddx[xn+1]  \Rightarrow {\left( {1 + x} \right)^n}\dfrac{d}{{dx}}\left( x \right) + x\dfrac{d}{{dx}}\left[ {{{\left( {1 + x} \right)}^n}} \right] = {C_0}\dfrac{d}{{dx}}\left[ {{x^1}} \right] + {C_1}\dfrac{d}{{dx}}\left[ {{x^2}} \right] + ..... \\\ .....{C_2}\dfrac{d}{{dx}}\left[ {{x^3}} \right] + {C_3}\dfrac{d}{{dx}}\left[ {{x^4}} \right] + ....... + {C_n}\dfrac{d}{{dx}}\left[ {{x^{n + 1}}} \right] \\\

As C0,C1,C2,.....,Cn{C_0},{C_1},{C_2},.....,{C_n} are constants.
Proceeding further we get

⇒(1+x)n×1+x×n(1+x)n−1=C0×1+C1×2x+C2×3x2+C3×4x3+.......+Cn×(n+1)xn ⇒(1+x)n+n(1+x)n−1x=C0+2C1x+3C2x2+4C3x3+.......+(n+1)Cnxn  \Rightarrow {\left( {1 + x} \right)^n} \times 1 + x \times n{\left( {1 + x} \right)^{n - 1}} = {C_0} \times 1 + {C_1} \times 2x + {C_2} \times 3{x^2} + {C_3} \times 4{x^3} + ....... + {C_n} \times \left( {n + 1} \right){x^n} \\\ \Rightarrow {\left( {1 + x} \right)^n} + n{\left( {1 + x} \right)^{n - 1}}x = {C_0} + 2{C_1}x + 3{C_2}{x^2} + 4{C_3}{x^3} + ....... + \left( {n + 1} \right){C_n}{x^n} \\\

Now, in order to remove x from the equation and to bring the equation similar to the result we will substitute x = -1 in the above equation.
⇒(1+(−1))n+n(1+(−1))n−1(−1)=C0+2C1(−1)+3C2(−1)2+4C3(−1)3+.......+(n+1)Cn(−1)n\Rightarrow {\left( {1 + \left( { - 1} \right)} \right)^n} + n{\left( {1 + \left( { - 1} \right)} \right)^{n - 1}}\left( { - 1} \right) = {C_0} + 2{C_1}\left( { - 1} \right) + 3{C_2}{\left( { - 1} \right)^2} + 4{C_3}{\left( { - 1} \right)^3} + ....... + \left( {n + 1} \right){C_n}{\left( { - 1} \right)^n}
Further evaluating the equation above we get:

⇒(1−1)n+n(1−1)n−1(−1)=C0−2C1+3C2−4C3+.......+(n+1)Cn(−1)n ⇒C0−2C1+3C2−4C3+.......+(n+1)Cn(−1)n=(0)n+n(0)n−1(−1) ⇒C0−2C1+3C2−4C3+.......+(−1)n(n+1)Cn=0  \Rightarrow {\left( {1 - 1} \right)^n} + n{\left( {1 - 1} \right)^{n - 1}}\left( { - 1} \right) = {C_0} - 2{C_1} + 3{C_2} - 4{C_3} + ....... + \left( {n + 1} \right){C_n}{\left( { - 1} \right)^n} \\\ \Rightarrow {C_0} - 2{C_1} + 3{C_2} - 4{C_3} + ....... + \left( {n + 1} \right){C_n}{\left( { - 1} \right)^n} = {\left( 0 \right)^n} + n{\left( 0 \right)^{n - 1}}\left( { - 1} \right) \\\ \Rightarrow {C_0} - 2{C_1} + 3{C_2} - 4{C_3} + ....... + {\left( { - 1} \right)^n}\left( {n + 1} \right){C_n} = 0 \\\

Hence the result is proved and we have C0−2C1+3C2−4C3+.......+(−1)n(n+1)Cn=0{C_0} - 2{C_1} + 3{C_2} - 4{C_3} + ....... + {\left( { - 1} \right)^n}\left( {n + 1} \right){C_n} = 0 .

Note: In order to solve such problems students should first try to visualize for some binomial expansion series. Students must not just start with the LHS in order to prove RHS. Students must remember the formula for binomial theorem for series expansion and methods of differentiation in order to solve such problems.