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

Question

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 (A) 110 (B) 111 (C) 120 (D) 121

Tardigrade Admin · Accepted Answer

k =1+2((10/11))+3((10/11))2+4((10/11))3+ ldots ldots ldots . . ∞ <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/b4548d9f5fc6ec0ac7e23fc176d27d6b-.png" /> ∴ k =(1/(1- x )2)=121

Q. If $(11)^{n} + 2 \cdot 1 1^{n - 1} \cdot 10 + 3 \cdot 1 1^{n - 2} \cdot 1 0^{2} + 4 \cdot 1 1^{n - 3} \cdot 1 0^{3} + \dots\dots\dots .\infty = k (11)^{n}$ then $k$ equals to

110

111

120

121

$k = 1 + 2 (\frac{10}{11}) + 3 (\frac{10}{11})^{2} + 4 (\frac{10}{11})^{3} + \dots\dots\dots ..\infty$

$∴ k = \frac{1}{( 1 - x ) ^{2}} = 121$

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

110

111

120

121

k=1+2(1110​)+3(1110​)2+4(1110​)3+………..∞ ∴k=(1−x)21​=121

Q. If $(11)^{n} + 2 \cdot 1 1^{n - 1} \cdot 10 + 3 \cdot 1 1^{n - 2} \cdot 1 0^{2} + 4 \cdot 1 1^{n - 3} \cdot 1 0^{3} + \dots\dots\dots .\infty = k (11)^{n}$ then $k$ equals to

$k = 1 + 2 (\frac{10}{11}) + 3 (\frac{10}{11})^{2} + 4 (\frac{10}{11})^{3} + \dots\dots\dots ..\infty$

$∴ k = \frac{1}{( 1 - x ) ^{2}} = 121$