Repeated
⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ a + b + c a 2 + b 2 + c 2 a 5 + b 5 + c 5 = 0 = 7 8 = − 1 3 6 5 0
Calculate a 7 + b 7 + c 7 .
This is part of the series: " It's easy, believe me! "
The answer is -745290.
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.
1 solution
You can also prove that 7 ( a 2 + b 2 + c 2 ) ( a 5 + b 5 + c 5 ) = 1 0 ( a 7 + b 7 + c 7 ) but coming with the idea itself in the first place is hard already.
Log in to reply
is this problem original ?
Log in to reply
Yes! Why would you think of that? Also I'm pretty sure that b 1 b 2 . . . b n − 1 b n ( x a 1 + y a 1 + z a 1 ) ( x a 2 + y a 2 + z a 2 ) . . . ( x a n − 1 + y a n − 1 + z a n − 1 ) ( x a n + y a n + z a n ) = a 1 a 2 . . . a n − 1 a n ( x b 1 + y b 1 + z b 1 ) ( x b 2 + y b 2 + z b 2 ) . . . ( x b n − 1 + y b n − 1 + z b n − 1 ) ( x b n + y b n + z b n ) if a + b + c = 0
How do you prove that?
Let P n = ⎩ ⎪ ⎨ ⎪ ⎧ 0 1 a n + b n + c n for n < 0 for n = 0 for n ≥ 1 , where n ∈ Z ; and ⎩ ⎪ ⎨ ⎪ ⎧ S 1 = a + b + c S 2 = a b + b c + c a S 3 = a b c . Then Newton's sums method is given by:
P n = S 1 P n − 1 − α 2 S 2 P n − 2 + α 3 S 3 P n − 3 , where α k = { k 1 for n = k for n > k
Therefore, we have:
\(\begin{array} {} P_1 = S_1P_0 = 0 \\ P_2 = S_1P_1 - 2S_2P_0 = 78 & \implies \color{blue} S_2 = - 39 \\ P_3 = 0 - S_1P_1 + 3S_3P_0 & \implies \color{blue} S_3 = \frac 13P_3 \\ P_4 = 39P_2 + S_3P_1 = 3042 \\ P_5 = 39P_3 + S_3P_2 = 195S_3 = -13650 & \implies \color{blue} S_3 = -70 \\ P_7 = 39P_5 -70P_4 = -745290 \end{array} \)
⟹ a 7 + b 7 + c 7 = P 7 = − 7 4 5 2 9 0