If (10)^9 + 2(11)^1(10)^8 + 3(11)^2(10)^7 + . | vedclass

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(A) Let the given series be $S = 10^9 + 2(11)^1(10)^8 + 3(11)^2(10)^7 + \dots + 10(11)^9 = k(10)^9$.Dividing both sides by $10^9$,we get:$k = 1 + 2\le

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(A) Let the given series be $S = 10^9 + 2(11)^1(10)^8 + 3(11)^2(10)^7 + \dots + 10(11)^9 = k(10)^9$.Dividing both sides by $10^9$,we get:$k = 1 + 2\left(\frac{11}{10}\right) + 3\left(\frac{11}{10}\right)^2 + \dots + 10\left(\frac{11}{10}\right)^9$  ......$(i)$Let $x = \frac{11}{10}$. Then $k = 1 + 2x + 3x^2 + \dots + 10x^9$.Multiply by $x$:$xk = x + 2x^2 + 3x^3 + \dots + 9x^9 + 10x^{10}$  ......$(ii)$Subtracting $(ii)$ from $(i)$:$k(1 - x) = 1 + x + x^2 + \dots + x^9 - 10x^{10}$$k(1 - x) = \frac{1(x^{10} - 1)}{x - 1} - 10x^{10}$Since $x = \frac{11}{10}$,$1 - x = -\frac{1}{10}$ and $x - 1 = \frac{1}{10}$.$k\left(-\frac{1}{10}\right) = \frac{(\frac{11}{10})^{10} - 1}{\frac{1}{10}} - 10\left(\frac{11}{10}\right)^{10}$$k\left(-\frac{1}{10}\right) = 10\left(\frac{11}{10}\right)^{10} - 10 - 10\left(\frac{11}{10}\right)^{10}$$k\left(-\frac{1}{10}\right) = -10$$k = 100$.

If $(10)^9 + 2(11)^1(10)^8 + 3(11)^2(10)^7 + ... + 10(11)^9 = k(10)^9$,then $k$ is equal to:

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