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  4. If (11) n +2 ⋅ 11 n -1 ⋅ 10+3 ⋅ 11 n -2 ⋅ 102+4 ⋅ 11 n -3 ⋅ 103+ ldots ldots ldots . ∞= k (11) n then k equals to

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Q. If $(11)^{ n }+2 \cdot 11^{ n -1} \cdot 10+3 \cdot 11^{ n -2} \cdot 10^2+4 \cdot 11^{ n -3} \cdot 10^3+\ldots \ldots \ldots . \infty= k (11)^{ n }$ then $k$ equals to

Sequences and Series

Solution:

$k =1+2\left(\frac{10}{11}\right)+3\left(\frac{10}{11}\right)^2+4\left(\frac{10}{11}\right)^3+\ldots \ldots \ldots . . \infty$
image
$\therefore k =\frac{1}{(1- x )^2}=121$