An algebra problem by A Former Brilliant Member

Algebra Level 4

If ( 10 ) 9 + 2 ( 11 ) 1 ⋅ ( 10 ) 8 + 3 ( 11 ) 2 ⋅ ( 10 ) 7 + . . . . + 10 ( 11 ) 9 = k ( 10 ) 9 (10)^9+2(11)^1\cdot(10)^8+3(11)^2\cdot(10)^7+....+10(11)^9=k(10)^9 ,bthen find the value of k k .


The answer is 100.0000.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

JEE-Mains 2014 (8/30) by Sandeep Bhardwaj, 24, India has the same problem.
Below is a copy of my solution in it.
k = ∑ n = 1 10 n ∗ 1. 1 n − 1 . a n A ⁣ − ⁣ G p r o g r e s s i o n . ∴ k = a − { a + ( n − 1 ) ∗ d } r n 1 − r + d ∗ r ( 1 − r n − 1 ) 1 − r 2 . O u r a = d = 1 , r = 1.1. S u b s t i t u t i n g w e g e t k = 100 . \displaystyle k=\sum_{n=1}^{10}n*1.1^{n-1}.~~~an~A\!-\!G~~progression.\\ ~~~~ \\ \therefore~~k=\dfrac{a-\{a+(n-1)*d\}r^n}{1-r} +\dfrac{d*r(1-r^{n-1}) }{1-r^2 }.\\ ~~\\ Our~a=d=1,~~r=1.1.\\ ~~~\\ Substituting~we~get~~\huge \color{#D61F06}{k = 100}.

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