Consider a recurrence relation an = -3an-1 + n for n = 1,2,3,4,… with initial conditions a1 = 3. Calculate a3.
Recurrence relation:
an=−3an−1+na_{n} = -3a_{n-1} + nan=−3an−1+n
Initial condition:
a1=3a_{1} = 3a1=3
Use recurrence relation for n=2n = 2n=2 :
a2=−3a1+2=−3∗3+2=−9+2=−7a_{2} = -3a_{1} + 2 = -3 * 3 + 2 =-9 + 2 = -7a2=−3a1+2=−3∗3+2=−9+2=−7
Use recurrence relation for n=3n = 3n=3 :
a3=−3a2+3=−3∗(−7)+3=21+3=24a_{3} = -3a_{2} + 3 = -3 * (-7) + 3 = 21 + 3 = 24a3=−3a2+3=−3∗(−7)+3=21+3=24
Answer: a3=24a_{3} = 24a3=24
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