Madness-4

Algebra Level 5

Let $S_n=(x^n+\frac{1}{x^n}).$

Now $(S_3+S_1)(S_7+S_1)(S_{11}+S_1)=(S_a+S_7)(S_b+S_7)(S_c+S_7)$

If the equation above is always true then find the value of $(a+b+c).$

This is one of my original Madness problems .

The answer is 9.

1 solution

Sanjeet Raria
Jul 31, 2015

Again applying the beautiful property, $\boxed{\large S_i S_j=S_{i+j}+S_{i-j}}$ $\Rightarrow (S_3+S_1)(S_7+S_1)(S_{11}+S_1)=(S_1S_2)(S_3S_4)(S_5S_6)$

$=(S_1S_6)(S_2S_5)(S_3S_4)=(S_5+S_7)(S_3+S_7)(S_1+S_7)$ $\Rightarrow a+b+c=5+3+1=\boxed9$

Mine solutions are not just same but identical to you:)

Aakash Khandelwal - 5 years, 10 months ago

Voted up. My method now is too long .

Niranjan Khanderia - 4 years, 11 months ago

Sir how could u prove this..................please send

Abhisek Mohanty - 4 years, 11 months ago

$(S_7+S_1)=(S_1+S_7)~ is~ all~ ready ~there.~so~a=1\\ \therefore~(S_3+S_1)(S_{11}+S_1)=(S_2*S_1)(S_6*S_5)\\ ~~~~~~~=(S_1*S_6)(S_2*S_5)=(S_5+S_7)(S_3+S_7)~~so~b=5,~c=3.\\ a+b+c+1+5+3=\Large~~~~\color{#D61F06}{9}.$

Niranjan Khanderia - 3 years, 4 months ago

Madness-4

This is one of my original Madness problems .

The answer is 9.

1 solution

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