142p23s11solution[1]

3 3 21 3 21 3 21 3 21 2 n

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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...
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This note was uploaded on 02/04/2014 for the course MAT 132 taught by Professor Poole during the Fall '08 term at SUNY Stony Brook.

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