NDA 2023
Exam NDA
Year 2023
Subject Mathematics
Chapter Binomial Theorem NDA 2023 · Mathematics · Binomial Theorem
Let (1+x)^n=1+T_1 x+T_2 x^2+T_3 x^3+ +T_n x^n What is T_1+2 T_2+3...
Let
( 1 + x ) n = 1 + T 1 x + T 2 x 2 + T 3 x 3 + … + T n x n (1+x)^n=1+T_1 x+T_2 x^2+T_3 x^3+\ldots+T_n x^n ( 1 + x ) n = 1 + T 1 x + T 2 x 2 + T 3 x 3 + … + T n x n
What is
T 1 + 2 T 2 + 3 T 3 + … + n T n T_1+2 T_2+3 T_3+\ldots+n T_n T 1 + 2 T 2 + 3 T 3 + … + n T n equal to?
Options A 0 B 1 C 2 n 2^{\mathrm{n}} 2 n D n 2 n − 1 n 2^{n-1} n 2 n − 1 Correct answer D. n 2 n − 1 n 2^{n-1} n 2 n − 1
Step-by-step solution
Given, ( 1 + x ) n = 1 + T 1 x + T 2 x 2 + … . + T n x n differentiating both side with respect to x , we get ⇒ n ( 1 + x ) n − 1 = T 1 + 2 T 2 x 1 + 2 T 3 x 2 + … . + n T n x n − 1 … ( i ) ⇒ n ( 1 + 1 ) n − 1 = T 1 + 2 T 2 + 3 T 3 + … . + n T n ⇒ T 1 + 2 T 2 + 3 T 3 + … . + n T n = n n − 1 {Putting x = 1 } \begin{aligned} & \text{ Given, }(1+\mathrm{x})^{\mathrm{n}}=1+\mathrm{T}_1 \mathrm{x}+\mathrm{T}_2 \mathrm{x}^2+\ldots.+\mathrm{T}_{\mathrm{n}} \mathrm{x}^{\mathrm{n}} \\ & \text{differentiating both side\text{ with } respect to } \mathrm{x} \text{, we get } \\ & \Rightarrow \mathrm{n}(1+\mathrm{x})^{\mathrm{n}-1}=\mathrm{T}_1+2 \mathrm{ T}_2 \mathrm{x}^1+2 \mathrm{ T}_3 \mathrm{x}^2+\ldots.+\mathrm{nT}_{\mathrm{n}} \mathrm{x}^{\mathrm{n}-1} \ldots(\mathrm{i}) \\ & \Rightarrow \mathrm{n}(1+1)^{\mathrm{n}-1}=\mathrm{T}_1+2 \mathrm{ T}_2+3 \mathrm{ T}_3+\ldots.+\mathrm{nT}_{\mathrm{n}} \\ & \left.\Rightarrow \mathrm{T}_1+2 \mathrm{ T}_2+3 \mathrm{ T}_3+\ldots.+\mathrm{nT}_{\mathrm{n}}=\mathrm{n}^{\mathrm{n}-1} \quad \text{ \{Putting } \mathrm{x}=1\right\}\end{aligned} Given, ( 1 + x ) n = 1 + T 1 x + T 2 x 2 + … . + T n x n differentiating both side with respect to x , we get ⇒ n ( 1 + x ) n − 1 = T 1 + 2 T 2 x 1 + 2 T 3 x 2 + … . + nT n x n − 1 … ( i ) ⇒ n ( 1 + 1 ) n − 1 = T 1 + 2 T 2 + 3 T 3 + … . + nT n ⇒ T 1 + 2 T 2 + 3 T 3 + … . + nT n = n n − 1 {Putting x = 1 }
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