If Δr =∣∣r−1(r−1)2(r−1)3n2n23n364n−23n2−3n∣∣. then ∑n=1nΔr is

Question

If   Δr​  =∣∣​r−1(r−1)2(r−1)3​n2n23n3​64n−23n2−3n​∣∣​. then  ∑n=1n​Δr​ is

Filo · Accepted Answer

We have,\begin{array}{l} \Delta x\left| { \begin{array} { *{ 20 }{ c } }{ r-1 } & n & 6 \ { { { \left( { r-1 } ight)  }^{ 2 } } } & { 2{ n^{ 2 } } } & { 4n-2 } \ { { { \left( { r-1 } ight)  }^{ 3 } } } & { 3{ n^{ 3 } } } & { 3{ n^{ 2 } }-3n } \end{array} } ight|  \ \sum _{ x=1 }^{ n }{ \Delta x\left| { \begin{array} { *{ 20 }{ c } }{ \sum  r-1 } & n & 6 \ { \sum { { \left( { r-1 } ight)  }^{ 2 } }  } & { 2{ n^{ 2 } } } & { 4n-2 } \ { \sum { { \left( { r-1 } ight)  }^{ 3 } }  } & { 3{ n^{ 3 } } } & { 3{ n^{ 2 } }-3n } \end{array} } ight|  }  \ =\left| { \begin{array} { *{ 20 }{ c } }{ \frac { { n\left( { r-1 } ight)  } }{ 2 }  } & n & 6 \ { \frac { { \left( { n-1 } ight) \left( { 2n-1 } ight)  } }{ 6 }  } & { 2{ n^{ 2 } } } & { 4n-2 } \ { \frac { { { { \left( { n-1 } ight)  }^{ 2 } }{ x^{ 2 } } } }{ 4 }  } & { 3{ n^{ 3 } } } & { 3{ n^{ 2 } }-3n } \end{array} } ight|  \ \sum _{ x=1 }^{ n }{ \Delta x=\frac { 1 }{ { 12 } } { x^{ 2 } }\left( { x-1 } ight) 	imes \left| { \begin{array} { *{ 20 }{ c } }6 & n & 6 \ { 2\left( { 2n-1 } ight)  } & { 2{ n^{ 2 } } } & { 2\left( { 2n-1 } ight)  } \ { 3n\left( { n-1 } ight)  } & { 3{ n^{ 3 } } } & { 3n\left( { n-1 } ight)  } \end{array} } ight|  }  \ =0 \end{array}Hence, this is the answer.

Question Text	If $Δ_{r} = ∣ ∣ r - 1 (r - 1)^{2} (r - 1)^{3} n 2 n^{2} 3 n^{3} 6 4 n - 2 3 n^{2} - 3 n ∣ ∣ .$ then $\sum_{n = 1}^{n} Δ_{r}$ is
Answer Type	Text solution:1
Upvotes	150

If $Δ_{r} = ∣ ∣ r - 1 (r - 1)^{2} (r - 1)^{3} n 2 n^{2} 3 n^{3} 6 4 n - 2 3 n^{2} - 3 n ∣ ∣ .$ then $\sum_{n = 1}^{n} Δ_{r}$ is

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If Δr​ =∣∣​r−1(r−1)2(r−1)3​n2n23n3​64n−23n2−3n​∣∣​. then ∑n=1n​Δr​ is

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Practice questions from similar books

If $Δ_{r} = ∣ ∣ r - 1 (r - 1)^{2} (r - 1)^{3} n 2 n^{2} 3 n^{3} 6 4 n - 2 3 n^{2} - 3 n ∣ ∣ .$ then $\sum_{n = 1}^{n} Δ_{r}$ is