analisis real 2

9
Induksi matematika 1) 1+ 2 + 3 + 4 ... + n = 1 2 n ( n+1 ) Penyelesaian Bukti : n = 1 → 1 = 1 2 x 1 ( 1 +1) = 1 2 ( 2) = 1 Bukti n = k → 1 + 2 + 3 + 4 ... + k = 1 2 k ( k +1) n = k + 1 → 1 + 2 + 3 + 4 + ... + k + k + 1 = 1 2 k +1 ( k+ 1+1 ) 1 2 k ( k +1 ) +k +1 = 1 2 ( k+ 1) ( k+2 ) 1 2 k 2 + 1 2 k +k +1 = 1 2 ( k 2 +2 k+ k+2 ) 1 2 k 2 + 3 2 k +1 = 1 2 k 2 + 3 2 k +1 ( terbukti ) 2) 1 + 4 + 9 + ... + n 2 = n ( n+1 ) ( 2 n +1) 6 Bukti n = 1 → 1 + 4 + 9 + ... + n 2 = n ( n+1 ) ( 2 n + 1) 6

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Induksi matematika1) 1+ 2 + 3 + 4 ... + n =Penyelesaian Bukti : n = 1 1 = == 1

Bukti n = k 1 + 2 + 3 + 4 ... + k = n = k + 1 1 + 2 + 3 + 4 + ... + k + k + 1 =

= = = ( terbukti )2) 1 + 4 + 9 + ... + Bukti n = 1 1 + 4 + 9 + ... + 1 1 1 1 ( terbukti )Bukti 2 n = k 1 1 + 4 + 9 + ... + = n = k + 1 1 + 4 + 9 + ... + = = x ( = = = + = + = (Terbukti)

3) 1,2 + 2,3 + 3,4 + ... + n ( n + 1 ) = Bukti n = 1 1 ( 1 + 1 ) = 2 = 2 = 2 ( terbukti )Bukti 2 n = kn = k + 1 1,2 + 2,3 + 3,4 + ... + k ( k + 1 ) + ( k + 1 ) ( k + 1 + 1 ) = , ( k + 1 ) ( k + 1 + 1 ) =

+ =

+ = + = = = = ( terbukti )

4) 1 + 2+ 3 + 2.3 + 3.4 + 3.4. 5 + ... n ( n + 1 ) ( n + 2 ) = n = 1 1 ( 1 + 1 ) ( 1 + 2 ) = 1.2.3= 6= 6 = 6 ( terbukti )Bukti 2 Di asumsikan n = k k ( k + 1 ) ( k + 2 ) = n= k + 1 1.2.3 + 2.3.4 + 3.4.5 + ... k( k + 1) (k + 2) + k + 1 ( k + 1 +1 ) (k + 1 + 2) = + k+1(k+2)(k+3) =+3k+2)(k+3) =+(+11k+6) == = = ( Terbukti )

5) 1+3+6+10+...+=n =1 =

1 = 1 = 1 ( Terbukti )

n = k 1 + 3 + 6+ 10 + .....+ = n = k 1 + 3 + 6+ 10 + .....+ + = + = = = = kedua ruas x6= (terbukti )

Barisan divergen dan konvergen 1. + + + ...

f(n)= f(x)= = dx= = ]- ]= tak hingga ( divergen)2. + + + ...

= dx= = ]= (divergen)

3. + + ....Sn = F(k) = f(x) = = dx= dx= du= x ]= x 1 = ( divergen)4. = + = + + ...

= dxU = = du= 2x = = 2xdx = du= Xdx = = . = ( divergen)

5. dx= U = x-3=du= dx= 2dx =2du= = 2]= = 2 ln = 2 (konvergen)

6. dx=