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Ronch [10]
3 years ago
11

Find the first 4 terms in the expansion of ( 1 + x 2 ) 8. Use your answer to find the value of ( 1.01 ) 8

Mathematics
1 answer:
anyanavicka [17]3 years ago
7 0

(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)

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4 0
3 years ago
Determine the sum of the arithmetic series: 5+18 +31 +44 + ... 161.
Finger [1]

Answer:

1079

Step-by-step explanation:

Hello,

18-5 = 13

31-18=13

44-31=13

161=5+13*12

So we need to compute

\displaystyle \sum_{k=0}^{k=12} \ {(5+13k)}\\\\=\sum_{k=0}^{k=12} \ {(5)} + 13\sum_{k=1}^{k=12} \ {(k)}\\\\=13*5+13*\dfrac{12*13}{2}\\\\=65+13*13*6\\\\=65+1014\\\\=1079

Thanks

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3 years ago
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Jet001 [13]

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Step-by-step explanation:

6 0
3 years ago
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nexus9112 [7]

Answer: You divide the top row by itself and you multiply the top by 2 each time.


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8 0
3 years ago
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Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
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