Halo teman-teman disini saya akan menyampaikan kumpulan soal matematika, pokok bahasan induksi matemtika, kami memberikan contoh dan jabawannya yaitu dengan pembuktian langkai satu dan dua. mari simak soal berikut ini ya .
Kumpulan Soal Induksi Matematika
#Soal ke 1
$2+6+12+20+... +$ $n(n+1$) = $ \frac {n(n+1)(n+2)}{3} $
Langkah 1
$p=1$
$n(n+1$) = $ \frac {n(n+1)(n+2)}{3} $
$1(1+1$) = $ \frac {1(1+1)(1+2)}{3} $
$2=2$
langkah 2
$p=k+1$
$2+6+12+20+$... + $n(n+1$) = $ \frac {n(n+1)(n+2)}{3} $
$2+6+12+20+$....+$k(k+1$)+(k+1)((k+1)+1)=$ \frac {(k+1)((k+1)+1)((k+1)+2)}{3} $
$ \frac {k(k+1)(k+2)}{3} $+ $(k+1)(k+2$)=$ \frac {(k+1)(k+2)(k+3)}{3} $
$ \frac {k(k+1)(k+2)}{3} $+ $\frac{3(k+1)(k+2)}{3}$= $ \frac {(k+1)(k+2)(k+3)}{3} $
$ \frac {k(k+1)(k+2)+3(k+1)(k+2)}{3} $ = $ \frac {(k+1)(k+2)(k+3)}{3} $
$ \frac {(k+1)(k+2)(k+3)}{3} $ = $ \frac {(k+1)(k+2)(k+3)}{3} $
(Terbukti)
#Soal ke 2
$2+7+12+17+.....$ $+ (5n-3) =$
$\frac {1}{2}n(5n-1)$
Langkah 1
$p=1$
$(5n-3) = \frac {n(5n-1)}{2} $
$5(1)-3) = \frac {1(5(1)-1)}{2} $
$ 5-3 = \frac {(5-1)}{2} $
2=2
$p=1$
$(5n-3) = \frac {n(5n-1)}{2} $
$5(1)-3) = \frac {1(5(1)-1)}{2} $
$ 5-3 = \frac {(5-1)}{2} $
2=2
langkah 2
$p=k+1$
$2+7+12+17+.....+5k-3$+$(5(k+1)-3$ =$ \frac {(k+1)(5(k+1)-1)}{2} $
$ \frac {1}{2}k(5k-1)+(5k+2) $ = $ ( $ $\frac {1}{2}k + $ $\frac {1}{2}) (5k+4) $
$ \frac {5}{2}k^2 $ - $\frac {1}{2}k + 5 k +2 $ = $ \frac {5}{2}k^2 + 2k $ + $ \frac {5}{2}k $ + 2
$ \frac {5}{2}k^2 $ + $\frac {9}{2}k +2 $ = $ \frac {5}{2}k^2 $ + $ \frac {9}{2}k $ + 2
# Soal ke 2:
$10+ 13+ 18+ 25+ 34+ 45+ 58...$ $+ n^2+9$ $=\frac{n^3+6n^2+53n}{6}$
Selanjutnya $n=1$
$n^2+9=1^2+9=10$ atau
$\frac{n^3+6n^2+53n}{6}$ = $\frac{1+6+53}{6}$
Selanjutnya $n=k$
$10+ 13+ 18+ 25+ 34+ 45+ 58...$ $+ k^2+9$ $=\frac{k^3+6k^2+53k}{6}$
Selanjutnya $n=k+1$
$10+ 13+ 18+ 25+ 34+ 45+ 58...$ $+ (k^2+9) +((k+1)^2+9)$ $=\frac{(k+1)^3+6(k+1)^2+53(k+1)}{6}$
Pembuktian:
$10+ 13+ 18+ 25+ 34+ 45+ 58...$ $+ (k^2+9) +((k+1)^2+9)$ $=\frac{(k+1)^3+6(k+1)^2+53(k+1)}{6}$
$\frac{k^3+6k^2+53k}{6} +((k+1)^2+9)$ $=\frac{(k+1)^3+6(k+1)^2+53(k+1)}{6}$
Sebelah kiri:
$=\frac{k^3+6k^2+53k}{6} +((k+1)^2+9)$
$= \frac{k^3+6k^2+53k}{6} +\frac{6k^2+12k^2+60}{6}$
$= \frac{k^3+12k^2+65k+60}{6}$
Sebelah kanan:
$=\frac{(k+1)^3+6(k+1)^2+53(k+1)}{6}$
$=$
#Soal ke 3
$3+7+11+15+19+$.... $ +(4n-1)= $ $2n^2$+n
Langkah 1
$n=1$
$4n-1=2n^2$+n
$4.1-1=2.1^2$+1
$4-1=2+1$
$3=3$
langkah 2
$n=k$
$3+7+11+15+19+$....+$(4k-1)$=$2k^2$+k
pembuktian n=k+1
$3+7+11+15+19+$....$+(4k-1)+(4(k+1)-1)$=$2(k+1)^2$+(k+1)
$2k^2$+k+(4(k+1)-1)=$2(k+1)^2$+(k+1)
$2k^2$+k+4k+4-1=$2(k+1)^2$+(k+1)
$2k^2$+5k+3=$2(k+1)^2$+(k+1)
$2k^2$+4k+k+2+1=$2(k+1)^2+(k+1)$
$2k^2$+4k+2+(k+1)$=$2(k+1)^2+(k+1)$
$2(k^2+2k+1)+(k+1)$=$2(k+1)^2+(k+1)$
$2(k+1)(k+1)+(k+1)$=$2(k+1)^2+(k+1)$
$2(k+1)^2+(k+1)$=$2(k+1)^2+(k+1)
#Soal ke 4
1+8+27+... +
$n^3$ = $ \frac {1}{4}
n^2 (n+1)^2 $
Langkah 1
$p=1$
$n^3$ = $ \frac {1}{4} n^2 (n+1)^2 $
$1^3$ = $ \frac {1}{4} 1^2 (1+1)^2 $
$1$ = $ \frac {1}{4} *1^2* 2^2 $
$1$ = $ \frac {4}{4}$
$1=1$
langkah 2
$p=k+1$
1+8+27+... + $k^3$+ $ (k+1)^3 $= $ \frac {1}{4} (k+1)^2 ((k+1)+1)^2 $
$ \frac {1}{4}k^2 (k+1)^2 (k+1)^3 $ = $\frac {1}{4} (k+1)^2 (k+2)^2 $
$ (k+1)^2 ( $ $\frac {1}{4} k^2 + (k+1) ) $= $\frac {1}{4} (k+1)^2 (k+2)^2 $
$ (k+1)^2 ( $ $ \frac {k^2+4(k+1)}{4} )$ = $\frac {1}{4} (k+1)^2 (k+2)^2 $
$ (k+1)^2 ( $ $ \frac {k^2+4k+4}{4}) $ = $\frac {1}{4} (k+1)^2 (k+2)^2 $
$ \frac {1}{4}(k+1)^2 (k^2+ 4k+4 )$ = $\frac {1}{4} (k+1)^2 (k+2)^2 $
$\frac {1}{4} (k+1)^2 (k+2)^2 $= $\frac {1}{4} (k+1)^2 (k+2)^2 $
#Soal ke 5
3+9+27+... +
$3^n$ = $ \frac {1}{2}
(3^{n+1}-3)$
Langkah 1
p=1
$3^n$ = $ \frac {1}{2} (3^{n+1}-3)$
$3^1$ = $ \frac {1}{2} (3^{1+1}-3)$
$3$ = $ \frac {1}{2} (3^2-3)$
$3$ = $ \frac {1}{2} (9-3)$
$3$ = $ \frac {6}{2} $
$3 = 3 $
langkah 2
$p=k+1$
3+9+27+... + $3^n$ = $ \frac {1}{2} (3^{n+1}-3)$
$ \frac {1}{2} (3^{k+1}-3)$ + $3^{k+1} = \frac {1}{2} (3^{k+2}-3)$
$ \frac {1}{2} 3^{k+1}-$ $ \frac {3}{2} $ = $3^{k+1}\ = \frac {1}{2} (3^{k+2}-3)$
$ 3^{k+1}$
$( \frac {3}{2}) $ - $ \frac {3}{2} =
\frac {1}{2}
(3^{k+2}-3)$
$ \frac {1}{2} (3^{k+2}-3)$ = $ \frac {1}{2} (3^{k+2}-3)$
$ \frac {1}{2} (3^{k+2}-3)$ = $ \frac {1}{2} (3^{k+2}-3)$
#Soal ke 6
$4+16+64... + $
$4^n$ = $ \frac {4}{3} (4^n-1)$
Langkah 1
$p=1$
$4^n$ = $ \frac {4}{3} (4^n-1)$
$4^1$ = $ \frac {4}{3} (4^1-1)$
$4$ = $ \frac {4}{3} (3)$
4=4
langkah 2
$p=k+1$
$4+16+64... + $ $4^k+4^{k+1}$ = $ \frac {(4(4^{(k+1)}-1)}{3}$
$\frac{4(4^k-1)}{3} + 4^{k+1} $= $ \frac {(4(4^{(k+1)}-1)}{3}$
$\frac{4(4^k-1)}{3} $+ $\frac{3}{3}$ $(4^{k+1}) $= $ \frac {(4(4^{(k+1)}-1)}{3}$
$\frac {4*4^k-4+3 (4^{k+1})}{3}$= $ \frac {(4*4^{(k+1)}-4)}{3}$
$\frac {4^{k+1}-4+3 (4^{k+1})}{3}$= $ \frac {(4^{k+2}-4)}{3}$
$\frac {4^{k+1}-4+4+3 (4^{k+1})}{3}$= $ \frac {(4^{k+2})}{3}$
$\frac {4^{k+1}+3 (4^{k+1})}{3}$= $ \frac {(4^{k+2})}{3}$
$ \frac {(4^{k+2})}{3}$ = $ \frac {(4^{k+2})}{3}$
#Soal ke 7
5+25+125+... + $5^n$= $ \frac {5}{4}(5^n-1) $
Langkah 1
$p=1$
$5^n$= $ \frac {5}{4}(5^n-1) $
$5^1$ = $ \frac {5}{4} (5^1-1) $
$5=5$
langkah 2
$p=k+1$
$5+25+125+...+$ $5^n$= $ \frac {5}{4}(5^n-1) $
$\frac {5}{4}(5^k-1)+5^k+1$=$ \frac {5}{4}(5^{(k+1)}-1) $
$\frac {5}{4}(5^k-1)+$ $\frac {45(5^k)}{4}$=$ \frac {5}{4}(5^{(k+1)}-1) $
$\frac {5}{4}(5^k-1)+$$4(5^k)$=$ \frac {5}{4}(5^{(k+1)}-1) $
$\frac {5}{4}(5(5^k)-1) =$ $\frac{5}{4}(5^{(k+1)}-1) $
$\frac {5}{4}(5^{(k+1)}-1) =$ $\frac{5}{4}(5^{(k+1)}-1) $
#Soal ke 8
$4+8+12+ ... +4n$
$=n(2n+2)$
Langkah 1
$p=1$
$4n = n(2n+2)$
$4(1) = 1(2(1)+2)$
$4 = 4$
langkah 2
$p=k+1$
$4+8+12+ ... +4n$ $=n(2n+2)$
$4+8+12+ ... +4k+4(k+1) $ $=(k+1)(2(k+1)+2)$
$k(2k+2) +4(k+1)$ $= (k+1) (2k+2+2)$
$$2k^2$ +2k+4k+4$ $= (k+1)(2k+4)$
$2k^2$+ 6k+ 4 $ $= (k+1)(2k+4)$
#Soal ke 9
6+12+24+... +
3.$2^n$ = 6 ($2^n$-1)
Langkah 1
$p=1$
3.$2^n$ = 6 ($2^n$-1)
3.$2^1$ = 6 ($2^1$-1)
$6 = 6$
langkah 2
$p=k+1 $
$6+12+24+... + $ $3.2^n+3.2^{k+1}$ $= 6 (2^{k+1} - 1^{})$ ,
$6 (2^k-1) + 3.2^{k+1}$ $= 6 (2^{k+1} - 1^{})$ ,
$6 .2^k - 6 + 3 (2^k.2^1)$ $= 6 (2^{k+1} - 1^{})$ ,
$6 .2^k - 6 + 6.2^k$ $= 6 (2^{k+1} - 1^{})$ ,
$12 .2^k - 6$ $= 6 (2^{k+1} - 1^{})$ ,
$6 (2^1.2^k - 1 )$ $= 6 (2^{k+1} - 1^{})$ ,
$6 (2^{k+1} - 1^{})$ $= 6 (2^{k+1} - 1^{})$ ,
Langkah 1
$p=1$
3.$2^n$ = 6 ($2^n$-1)
3.$2^1$ = 6 ($2^1$-1)
$6 = 6$
langkah 2
$p=k+1 $
$6+12+24+... + $ $3.2^n+3.2^{k+1}$ $= 6 (2^{k+1} - 1^{})$ ,
$6 (2^k-1) + 3.2^{k+1}$ $= 6 (2^{k+1} - 1^{})$ ,
$6 .2^k - 6 + 3 (2^k.2^1)$ $= 6 (2^{k+1} - 1^{})$ ,
$6 .2^k - 6 + 6.2^k$ $= 6 (2^{k+1} - 1^{})$ ,
$12 .2^k - 6$ $= 6 (2^{k+1} - 1^{})$ ,
$6 (2^1.2^k - 1 )$ $= 6 (2^{k+1} - 1^{})$ ,
$6 (2^{k+1} - 1^{})$ $= 6 (2^{k+1} - 1^{})$ ,
#Soal ke 10
$6+18+54+162+486+1458... +$
$2*3^n=$ $ 3(3^n-1) $
Langkah 1
$p=1$
$2*3^n$ = $ 3(3^n-1) $
$2*3^1$= $ 3(3^1-1) $
$6$= $ 3(2) $
$6=6$
langkah 2
$p=k+1$
$6+18+54+162+486+1458... + $ $2*3^k$ + $ 2(3^{k+1}) $= $ 3(3^{k+1}-1) $
$ 3 (3^k-1)$+ $2*3^{k+1} =$ $ 3(3^{k+1}-1) $
$p=1$
$2*3^n$ = $ 3(3^n-1) $
$2*3^1$= $ 3(3^1-1) $
$6$= $ 3(2) $
$6=6$
langkah 2
$p=k+1$
$6+18+54+162+486+1458... + $ $2*3^k$ + $ 2(3^{k+1}) $= $ 3(3^{k+1}-1) $
$ 3 (3^k-1)$+ $2*3^{k+1} =$ $ 3(3^{k+1}-1) $
$ 3^{k+1} -3 + 2 * 3^{k+1} =$ $ 3^{k+2}-3$
$ 3^{k+1} + 2 * 3^{k+1}= $ $ 3^{k+2}$
$ (1+2) 3^{k+1} =$ $ 3^{k+2}$
$ 3* 3^{k+1} =$ $ 3^{k+2}$
$ 3^1 * 3^{k+1} =$ $ 3^{k+2}$
$ 3^{k+2}=$ $ 3^{k+2}$
$ k+2=$ $ k+2$
$ 3* 3^{k+1} =$ $ 3^{k+2}$
$ 3^1 * 3^{k+1} =$ $ 3^{k+2}$
$ 3^{k+2}=$ $ 3^{k+2}$
$ k+2=$ $ k+2$
