Is this even possible?
Given that a + b = 3 and a 2 + b 2 = 7 , find the value of
a 1 0 + 2 a 9 b + 3 a 8 b 2 + 4 a 7 b 3 + 5 a 6 b 4 + 6 a 5 b 5 + 5 a 4 b 6 + 4 a 3 b 7 + 3 a 2 b 8 + 2 a b 9 + b 1 0 .
The answer is 20736.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
2 solutions
One could also use the Pascal's triangle to determine the expansion of ( a + b ) 1 0 = 3 1 0 and accordingly subtract the binomials (after figuring out that a b = 1 ) a 8 + b 8 , a 6 + b 6 , a 4 + b 4 , a 2 + b 2 (with some coefficients accordingly) and a constant number, which would let us avoid calculating a 1 0 + b 1 0 . And another method would be using brute-force - solving the system of equations { a + b = 3 a 2 + b 2 = 7 and putting the values of a , b into the expression. That would be tedious, though.
i did a calculation error
We have to find ( a − b a 6 − b 6 ) 2 and we know that a and b are roots of t 2 − 3 t + 1 , so ( a − b ) 2 equals the discriminant of t 2 − 3 t + 1 , i.e. 1 3 , while ( a 6 − b 6 ) 2 equals the discriminant of the minimal polynomial of a 6 . Since t 2 = 3 t − 1 implies t 3 = 3 t 2 − t = 8 t − 3 and t 6 = 6 4 t 2 − 4 8 t + 9 = 1 4 4 t − 5 5 , the ratio between the two discriminants is just 1 4 4 2 = 2 0 7 3 6 .
I did not know that could be factored! I just Newton's Sums bashed out all a n + b n up to n = 1 0 .
The value can be express by the formula: S = ( a 1 0 + b 1 0 + 2 ( a 8 + b 8 ) + 3 ( a 6 + b 6 ) + 4 ( a 4 + b 4 ) + 5 ( a 2 + b 2 ) + 6 Because, the product ab=1 ⇒ a 3 + b 3 = 1 8 , a 4 + b 4 = 4 7 , a 8 + b 8 = 4 7 2 − 2 , a 6 + b 6 = 1 8 2 − 2 , a 5 + b 5 = ( a 2 + b 2 ) ( a 3 + b 3 ) − 3 = 7 ∗ 1 8 − 3 = 1 2 3 , a 1 0 + b 1 0 = 1 2 3 2 − 2 S = 2 0 7 3 6