(1+x^2)^8
=(1+8x^2+8*7/(1*2)x^4+8*7*6/(1*2*3)x^6+8*7*6*5/(1*2*3*4)x^8+....)
=1+8x^2+28x^4+56x^6+70x^8+....)
For x<1, higher power terms diminish in value, hence we can approximate powers of numbers.
1.01=(1+0.1^2) => x=0.1 in the above expansion
(1.01)^8
=1+8(0.1^2)+28(0.1^4)+56(0.1^6) [ limited to four terms, as requested]
=1+0.08+0.0028+0.000056 (+0.00000070)
=1.082856 (approximately)
Answer:
A: A high unemployment rate
Step-by-step explanation:
