Question

If 6x⁵+5x³ + x² + 4x -1=a(x-1)⁵ +b(x-1)⁴+ c(x-1)³+d(x-1)²+e(x-1)+f . Find a +2b +3c+ 4d + 5e + 6f.

Answer 1

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Answer 2

Given:

6x⁵+5x³ + x² + 4x -1 = a(x-1)⁵ + b(x-1)⁴+ c(x-1)³+ d(x-1)²+ e(x-1) + f

To Find:

The value of a + 2b + 3c + 4d + 5e + 6f

Solution:

Let us re-write the L.H.S as

6(x–1 +1)⁵+ 5(x–1 +1)³ + (x–1 +1)²+ 4(x–1 +1) – 1

Now we put y = x–1 to the equation and we get the following equation,

= 6(y+1)⁵ + 5(y+1)³+ (y+1)² + 4(y+1) – 1

The expansion of the above equation gives,

= (6y⁵+30y⁴+60y³+60y²+30y+6) + (5y³+15y²+15y+5)+ (y²+2y+1) + (4y+4) – 1

= 6y⁵+30y⁴+65y³+76y²+51y+15

Now on replacing y = x–1 in the above equation just formed we have,

6(x–1)⁵ +30(x–1)⁴ + 65(x–1)³ + 76(x–1)² + 51(x–1) + 15

Hence, from this equation we can get the corresponding values of the corresponding coefficients of the equation as:

6(x–1)⁵ +30(x–1)⁴ + 65(x–1)³ + 76(x–1)² + 51(x–1) + 15 = a(x-1)⁵ + b(x-1)⁴ + c(x-1)³+ d(x-1)²+ e(x-1) + f

a is equal to 6

b is equal to 30

c is equal to 65

d is equal to 76

e is equal to 51

f is equal to 15

Finally, a + 2b + 3c + 4d + 5e + 6f is

= 6 + 2(30) + 3(65) + 4(76) + 5(51) + 6(15)

= 6 + 60 + 195 + 304 + 255 + 90

= (6+304) + (195+255) + (60+90)

= 310 + 450 + 150

= 310 + 600

= 910.

Hence, a+2b+3c+4d+5e+6f = 910.

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