142p23s11solution[1]

# 3 3 21 3 21 3 21 3 21 2 n

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: s ( 3θ ) + 3θ ) ) 1 3 1 2 π 3 0 =π 3 x = r cos θ = 2sin ( 3θ ) cos θ and y = r sin θ = 2sin ( 3θ ) sin θ = 6 cos ( 3θ ) cos θ − 2sin ( 3θ ) sin θ and dy dx = dy dθ dy dθ = 6 cos ( 3θ ) sin θ + 2sin ( 3θ ) cos θ ⋅ dθ dx 6 cos ( 3θ ) sin θ + 2sin ( 3θ ) cos θ dy = dx θ = π6 6 cos ( 3θ ) cos θ − 2sin ( 3θ ) sin θ (c) 1 2 5π 18 ∫ ( 4sin ( 3θ ) − 1) dθ π 18 2 =− 3 θ =π 6 lim sn = lim 3 + ( (c) 2.18 (a) (b) 2.17 3 4+ n 4n 2 + 3n 4n 2 + 3n n12 lim sn = lim 2 = lim 2 ⋅ 1 = lim =2 n →∞ n →∞ 2n + 3n + 1 n →∞ 2 n + 3n + 1 n →∞ 2 + 3 + 1 n n2 n2 lim (a) 3 − 3 + 232 − 233 + 234 − ... = 3 + 3 ( −21 ) + 3 ( −21 ) + 3 ( −21 ) + 3 ( −21 ) + ... 2 n →∞ n →∞ n2 n →∞ 2 n L. R . ( n n ) = 3 + lim ( −1) n →∞ 2n n n →∞ 2 ln 2 n L. R. = lim = lim n →∞ n =3 2 2n ( ln 2 ) 2 =0 2 ∞ = 3∑ ( −21 ) = 21 ∑ 3a 3 n n =0 (b) −1) 2n 1 − ( −21 ) 3 4 =2 = 3a 2 + 3a 4 + 3a 6 + ... + 3a 42 = 3a 2 (1 + a 2 + a 4...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online