Expand (1+2x/3)8 in ascending powers of x up to the fourth term. Hence use your expansion to evaluate (0.98)8 to three significant figures.

Question 1

Expand (1+^2x/₃)⁸ in ascending powers of x up to the fourth term. Hence use your expansion to evaluate (0.98)⁸ to three significant figures.

Question 2

1 + 8(^2x/₃) + 28(^2x/₃)2 + 56(^2x/₃)³
= 1 + ^16x/₃ + 112x²/9 + 448x³/27
(1 + ^2x/₃) = [ 1 + (-0.02)]
^2x/₃ = -0.02
x = -0.03
0.988
1 + ¹⁶/₃(-0.03) + ¹¹²/₉(-0.03)²
+ ⁴⁴⁸/₂₇( -0.02)³
1 - 0.16 + 0.0112 - 0.000448
= 0.851

anonymous · Answer 1 · 2021-06-30T08:17:08+0000

1 + 8(^2x/₃) + 28(^2x/₃)2 + 56(^2x/₃)³
= 1 + ^16x/₃ + 112x²/9 + 448x³/27
(1 + ^2x/₃) = [ 1 + (-0.02)]
^2x/₃ = -0.02
x = -0.03
0.988
1 + ¹⁶/₃(-0.03) + ¹¹²/₉(-0.03)²
+ ⁴⁴⁸/₂₇( -0.02)³
1 - 0.16 + 0.0112 - 0.000448
= 0.851

Expand (1+2x/3)8 in ascending powers of x up to the fourth term. Hence use your expansion to evaluate (0.98)8 to three significant figures.

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