# Given #f(x) = x^4+ax^3+bx^2+cx+d# with #f(1) = 1#, #f(2) = 5#, #f(3) = 19# and #f(6) = 665#, what is the value of #f(4)+f(5)# ?

##### 2 Answers

#### Answer:

Solution

#### Explanation:

Given

solving gives the solution as

Substitution will give you

#### Answer:

#### Explanation:

Since we're dealing with

Define:

#a_n = n^4 - f(n) = -(an^3+bn^2+cn+d)#

Since this is a cubic, then starting with

#{ (a_1 = 1^4 - f(1) = 1-1 = 0), (a_2 = 2^4 - f(2) = 16 - 5 = 11), (a_3 = 3^4 - f(3) = 81 - 19 = 62), (a_4 = 4^4 - f(4) = 256 - f(4)), (a_5 = 5^4 - f(5) = 625 - f(5)), (a_6 = 6^4 - f(6) = 1296 - 665 = 631) :}#

So our initial sequence is:

#0, 11, 62, a_4, a_5, 631#

The sequence of differences between successive pairs of terms is:

#11, 51, (a_4 - 62), (a_5 - a_4), (631 - a_5)#

The sequence of differences of those differences is:

#40, (a_4 - 113), (a_5 - 2a_4 + 62), (a_4 - 2a_5 + 631)#

The sequence of differences of those differences is:

#(a_4 - 153), (a_5 - 3a_4 + 175), (3a_4 - 3a_5 + 569)#

The sequence of differences of those differences is:

#(a_5 - 4a_4 + 328), (6a_4-4a_5+394)#

So:

#{ (a_5 - 4a_4 + 328 = 0), (6a_4 - 4a_5 + 394 = 0) :}#

Dividing the second of these equations by

#3a_4-2a_5+197 = 0#

Adding this to the first equation, we get:

#-a_4-a_5+525=0#

Hence:

#f(4)+f(5) = (256-a_4)+(625-a_5) = -a_4-a_5+525+356 = 356#