Let -r - r-1 n 12 r-12 2n2 8n-4 r-13 3n3 6n2-6n- Then the value of -r-1n-r is-

Since C_1 has variable terms and nbsp;C_2 and nbsp;C_3 are constant, summation runs only on nbsp;C_1. ∴ nbsp; nbsp; ∑_r=1^nΔ_r = nbsp;∑_r=1^n( r 1) ...

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Standard XII

Mathematics

Infinite GP

Let Δr = r-1 ...

Question

Let $Δ_{r} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} r - 1 & n & 12 (r - 1)^{2} & 2 n^{2} & 8 n - 4 (r - 1)^{3} & 3 n^{3} & 6 n^{2} - 6 n \end{matrix} ∣ ∣ ∣ ∣ ∣$ . Then the value of $n \sum r = 1 Δ_{r}$ is:

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Solution

The correct option is B $0$
Since $C_{1}$ has variable terms and $C_{2}$ and $C_{3}$ are constant, summation runs only on $C_{1}$ .
$∴ n \sum r = 1 Δ_{r} = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} n \sum r = 1 (r - 1) & n & 12 n \sum r = 1 (r - 1)^{2} & 2 n^{2} & 8 n - 4 n \sum r = 1 (r - 1)^{3} & 3 n^{3} & 6 n^{2} - 6 n \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$
$= ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{1}{2} (n - 1) n & n & 12 \frac{1}{6} (n - 1) n (2 n - 1) & 2 n^{2} & 8 n - 4 \frac{1}{4} (n - 1)^{2} n^{2} & 3 n^{3} & 6 n^{2} - 6 n \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$

Taking $\frac{1}{12} n (n - 1)$ common from $C_{1}$ and $2$ from $C_{3}$ , we get
$n \sum r = 1 Δ_{r} = \frac{1}{6} n (n - 1) \times ∣ ∣ ∣ ∣ ∣ \begin{matrix} 6 & n & 6 2 (2 n - 1) & 2 n^{2} & 2 (2 n - 1) 3 n (n - 1) & 3 n^{3} & 3 n (n - 1) \end{matrix} ∣ ∣ ∣ ∣ ∣$
$\Rightarrow \sum Δ_{r} = 0$ , since $C_{1}$ and $C_{3}$ are identical.

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Let Δr=∣∣ ∣ ∣∣r−1n12(r−1)22n28n−4(r−1)33n36n2−6n∣∣ ∣ ∣∣. Then the value of n∑r=1Δr is:

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