KVPY #12

Algebra Level 4

6 x 5 + 5 x 3 + x 2 + 4 x − 1 = a ( x − 1 ) 5 + b ( x − 1 ) 4 + c ( x − 1 ) 3 + d ( x − 1 ) 2 + e ( x − 1 ) + f 6x^{5} + 5x^{3} + x^{2} + 4x-1 = a(x-1)^{5} + b(x-1)^{4}+ c(x-1)^{3}+ d(x-1)^{2} + e(x-1) +f

The equation above is an identity for constants a , b , c , d , e , f a,b,c,d,e,f .

What is the value of a + 2 b + 3 c + 4 d + 5 e + 6 f ? a + 2b + 3c + 4d + 5e + 6f?

For more KVPY questions, try my set


The answer is 910.

Rahil Sehgal by Rahil Sehgal , 20, India

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2 solutions

Rahil Sehgal
Mar 28, 2017

4 Helpful 0 Interesting 0 Brilliant 0 Confused
Rab Gani
Apr 24, 2017

Comparing coefficients of x^5, and x^4,resp, we get a=6, 6(5)(-1) + b =0, b=30 6(x – 1)^5 + 30(x – 1)^4 +c(x – 1)^3 +d(x – 1)^2 +e(x – 1) +f = 6x^5 + 5x^3 +x^2 +4x – 1 Comparing coefficients of x^3, we get : 6(10) – 30(4) – c = 5, c=65. Comparing coefficients of x^2, we get : 6(-10) + 30(6) – (65)(3) – d = 1, c=76. Putting x= 1, then f = 15. Putting x= 0, then -6+30-65+76-e+15=-1. e=51 So a+2b+3c+4d+5e+6f = 910

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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