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given that ax+by=3,ax 2 +by 2 =7,ax 3 +by 3 =16,ax 4 +by 4 =42.Find ax 5 +by 5 value.a,b,x,y are positive integers

given that ax+by=3,ax2+by2=7,ax3+by3=16,ax4+by4=42.Find ax5+by5 value.a,b,x,y are positive integers

Grade:9

1 Answers

SHAIK AASIF AHAMED
askIITians Faculty 74 Points
9 years ago
Hello student,
Please find the answer to your question below
We have ax3 + by3 = 16,
so (ax3+ by3)(x + y) = 16(x + y)
and thusax4+ by4 + xy(ax2+ by2) = 16(x + y)
42 + 7xy = 16(x + y)
From ax2+ by2 = 7, we have
(ax2+ by2)(x + y) = 7(x + y)
so ax3+ by3 + xy(ax2+ by2) = 7(x + y).
This implies to
16 + 3xy = 7(x + y)
We can now solve for x + y and xy from (1) and (2) to find x + y=-14 and xy=38
Thus we have
(ax4+ by4)(x + y) = 42(x + y)
and so ax5+ by5 + xy(ax3 + by3) = 42(x + y).
Finally, it follows that
ax5+ by5=42(x+y)-16xy=20

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