Prove The Following Equation C0 2C1 + 3C2 4C3 + + Left ... - Vedantu

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AnswerQuestion Answers for Class 12Class 12 BiologyClass 12 ChemistryClass 12 EnglishClass 12 MathsClass 12 PhysicsClass 12 Social ScienceClass 12 Business StudiesClass 12 EconomicsQuestion Answers for Class 11Class 11 EconomicsClass 11 Computer ScienceClass 11 BiologyClass 11 ChemistryClass 11 EnglishClass 11 MathsClass 11 PhysicsClass 11 Social ScienceClass 11 AccountancyClass 11 Business StudiesQuestion Answers for Class 10Class 10 ScienceClass 10 EnglishClass 10 MathsClass 10 Social ScienceClass 10 General KnowledgeQuestion Answers for Class 9Class 9 General KnowledgeClass 9 ScienceClass 9 EnglishClass 9 MathsClass 9 Social ScienceQuestion Answers for Class 8Class 8 ScienceClass 8 EnglishClass 8 MathsClass 8 Social ScienceQuestion Answers for Class 7Class 7 ScienceClass 7 EnglishClass 7 MathsClass 7 Social ScienceQuestion Answers for Class 6Class 6 ScienceClass 6 EnglishClass 6 MathsClass 6 Social ScienceQuestion Answers for Class 5Class 5 ScienceClass 5 EnglishClass 5 MathsClass 5 Social ScienceQuestion Answers for Class 4Class 4 ScienceClass 4 EnglishClass 4 MathsProve the following equation:${C_0} - 2{C_1} + 3{C_2} - 4{C_3} + ....... + {\left( { - 1} \right)^n}\left( {n + 1} \right){C_n} = 0$ AnswerVerified603.6k+ viewsHint: 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 ${C_0} - 2{C_1} + 3{C_2} - 4{C_3} + ....... + {\left( { - 1} \right)^n}\left( {n + 1} \right){C_n} = 0$ As we know the general formula for the binomial expansion of the series is given by:\[{\left( {1 + x} \right)^n} = {C_0}{x^0} + {C_1}{x^1} + {C_2}{x^2} + {C_3}{x^3} + ....... + {C_n}{x^n}\] Multiplying both sides of the above equation by x we get:\[ \Rightarrow x \times {\left( {1 + x} \right)^n} = x \times \left( {{C_0}{x^0} + {C_1}{x^1} + {C_2}{x^2} + {C_3}{x^3} + ....... + {C_n}{x^n}} \right) \\ \Rightarrow {\left( {1 + x} \right)^n} \cdot x = {C_0}{x^1} + {C_1}{x^2} + {C_2}{x^3} + {C_3}{x^4} + ....... + {C_n}{x^{n + 1}} \\ \] Now in order bring the equation something in form of the result let us differentiate the whole equation with respect to x\[ \Rightarrow \dfrac{d}{{dx}}\left[ {{{\left( {1 + x} \right)}^n} \cdot x} \right] = \dfrac{d}{{dx}}\left[ {{C_0}{x^1} + {C_1}{x^2} + {C_2}{x^3} + {C_3}{x^4} + ....... + {C_n}{x^{n + 1}}} \right]\] Now let us open the brackets in order to differentiate the whole term\[ \Rightarrow \dfrac{d}{{dx}}\left[ {{{\left( {1 + x} \right)}^n} \cdot x} \right] = \dfrac{d}{{dx}}\left[ {{C_0}{x^1}} \right] + \dfrac{d}{{dx}}\left[ {{C_1}{x^2}} \right] + \dfrac{d}{{dx}}\left[ {{C_2}{x^3}} \right] + \dfrac{d}{{dx}}\left[ {{C_3}{x^4}} \right] + ....... + \dfrac{d}{{dx}}\left[ {{C_n}{x^{n + 1}}} \right]\] As we know the formulas for differentiation are:\[\dfrac{d}{{dx}}\left( {{x^k}} \right) = k{x^{k - 1}}{\text{ and }}\dfrac{d}{{dx}}\left( {p.q} \right) = p\dfrac{d}{{dx}}\left( q \right) + q\dfrac{d}{{dx}}\left( p \right)\] Using the above formulas we differentiate the terms in the equation:\[ \Rightarrow {\left( {1 + x} \right)^n}\dfrac{d}{{dx}}\left( x \right) + x\dfrac{d}{{dx}}\left[ {{{\left( {1 + x} \right)}^n}} \right] = {C_0}\dfrac{d}{{dx}}\left[ {{x^1}} \right] + {C_1}\dfrac{d}{{dx}}\left[ {{x^2}} \right] + ..... \\ .....{C_2}\dfrac{d}{{dx}}\left[ {{x^3}} \right] + {C_3}\dfrac{d}{{dx}}\left[ {{x^4}} \right] + ....... + {C_n}\dfrac{d}{{dx}}\left[ {{x^{n + 1}}} \right] \\ \] As \[{C_0},{C_1},{C_2},.....,{C_n}\] are constants.Proceeding further we get\[ \Rightarrow {\left( {1 + x} \right)^n} \times 1 + x \times n{\left( {1 + x} \right)^{n - 1}} = {C_0} \times 1 + {C_1} \times 2x + {C_2} \times 3{x^2} + {C_3} \times 4{x^3} + ....... + {C_n} \times \left( {n + 1} \right){x^n} \\ \Rightarrow {\left( {1 + x} \right)^n} + n{\left( {1 + x} \right)^{n - 1}}x = {C_0} + 2{C_1}x + 3{C_2}{x^2} + 4{C_3}{x^3} + ....... + \left( {n + 1} \right){C_n}{x^n} \\ \] 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.\[ \Rightarrow {\left( {1 + \left( { - 1} \right)} \right)^n} + n{\left( {1 + \left( { - 1} \right)} \right)^{n - 1}}\left( { - 1} \right) = {C_0} + 2{C_1}\left( { - 1} \right) + 3{C_2}{\left( { - 1} \right)^2} + 4{C_3}{\left( { - 1} \right)^3} + ....... + \left( {n + 1} \right){C_n}{\left( { - 1} \right)^n}\] Further evaluating the equation above we get:\[ \Rightarrow {\left( {1 - 1} \right)^n} + n{\left( {1 - 1} \right)^{n - 1}}\left( { - 1} \right) = {C_0} - 2{C_1} + 3{C_2} - 4{C_3} + ....... + \left( {n + 1} \right){C_n}{\left( { - 1} \right)^n} \\ \Rightarrow {C_0} - 2{C_1} + 3{C_2} - 4{C_3} + ....... + \left( {n + 1} \right){C_n}{\left( { - 1} \right)^n} = {\left( 0 \right)^n} + n{\left( 0 \right)^{n - 1}}\left( { - 1} \right) \\ \Rightarrow {C_0} - 2{C_1} + 3{C_2} - 4{C_3} + ....... + {\left( { - 1} \right)^n}\left( {n + 1} \right){C_n} = 0 \\ \] Hence the result is proved and we have \[{C_0} - 2{C_1} + 3{C_2} - 4{C_3} + ....... + {\left( { - 1} \right)^n}\left( {n + 1} \right){C_n} = 0\] .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.Recently Updated PagesBasicity of sulphurous acid and sulphuric acid areMaster Class 12 English: Engaging Questions & Answers for SuccessMaster Class 12 Social Science: Engaging Questions & Answers for SuccessMaster Class 12 Maths: Engaging Questions & Answers for SuccessMaster Class 12 Economics: Engaging Questions & Answers for SuccessMaster Class 12 Physics: Engaging Questions & Answers for SuccessBasicity of sulphurous acid and sulphuric acid areMaster Class 12 English: Engaging Questions & Answers for SuccessMaster Class 12 Social Science: Engaging Questions & Answers for SuccessMaster Class 12 Maths: Engaging Questions & Answers for SuccessMaster Class 12 Economics: Engaging Questions & Answers for SuccessMaster Class 12 Physics: Engaging Questions & Answers for Success

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