Dengan Induksi matemtika buktikan bahwa 2+7+12+17+...+ (5n-3) = $$\frac{1}{2}$$ n (5n-1)

D. 2 ; 2+7+12+17+...+ (5k-3= $$\frac{1}{2}k\ \left(5k-1\right)\ ;\ \frac{5k^2+9k+4}{2}$$

A. 2 ; 2+7+12+17+...+ (5k-3) = $$\frac{1}{2}k\ \left(5k-1\right)\ ;\ \frac{5k^2+3k+2}{4+2}$$

B. 2 ; 2+7+12+17+...+ (4k-2k)= $$\frac{1}{2}k\ \left(5k+2\right)\ ;\ \frac{5k^{ }+6k+2}{2}$$

C. 2 ; 2+7+12+17+...+ (k-3= $$\frac{1}{4}k\ \left(4k-1\right)\ ;\ \frac{2k^2+8k+4}{2}$$

E. ; 2+7+12+17+...+ (10k-3= $$\frac{1}{8}k\ \left(14k-12\right)\ ;\ \frac{6k^2+9k+4}{2}$$

Buktikanlah bahwa $$n^3+2n$$ habis dibagi 3 untuk n bilangan asli

C. 3 ; $$k^3$$ + 2k ; $$\left(k^3+2k\right)\ +\ 3\ \left(k^2+k+1\right)$$

A. 6 ; $$k^3$$ + 2k ; $$\left(k^3+3k\right)\ +\ 6\ \left(k^2+3k+1\right)$$

B. - 3 ; $$k^3$$ + 2k ; $$\left(k^{ }+2k\right)\ +\ 3\ \left(k^{ }+k+1\right)$$

D. 12 ; $$k^3$$ + 2k ; $$\left(k^3-2k\right)\ +\ 2\ \left(k^2-k-2\right)$$

E. 3 ; $$k^3$$ + 2k ; $$\left(3k^{ }+3k^3\right)\ +\ 3\ \left(3k^2+9k+1\right)$$

Dengan Induksi matematika buktikanlah bahwa 2+7+12+17+...+(5n-3) = $$\frac{1}{2}$$ 𝑛 (5n − 1)

E. 2 ; 2+7+12+17+... (5k-3) $$=\frac{1}{2}(k)\left(5k-1\right)\ ;\ \frac{5k^2\ +9k+4}{2}$$

A. . 1 ; 2+7+12+17+... (5k-3) $$=\frac{1}{2}(k)\ \left(5k-1\right)\ ;\ \frac{5k^2-8k\ +4}{2}$$

B. 2 ; 2+7+12+17+... (5k-3) $$=\frac{1}{2}(k)(8-1)\ ;\ \frac{7k^2+9k+4}{8}^{ }$$

C. 3; 2+7+12+17+... (5k-3) $$=\frac{1}{2}(k)\left(5k-1\right);\frac{5k^2+9k+4}{-2}$$

D. -2 ; 2+7+12+17+... (5k-3) $$\frac{1}{2}\left(k\right)\ \left(5k-1\right)\ ;\ \frac{4k-5k+4\ }{6}$$

Dengan Induksi Matematika buktikanlah bahwa 1²+2²+3³+..+n² = n $$\frac{\left(n+1\right)\ \left(2n+1\right)}{6}$$

D. 1; 1²+2²+3³+k ²= $$\frac{\text{k(k+1) (2k+1)}}{6};\ \frac{\left(k+1\right)\left(k+2\right)\left(2k+3\right)}{6}$$

A . 1; 1²+2²+3³+k ²= $$\frac{\text{k(k+1) (k+1)}}{6};\ \frac{\left(k+1\right)\left(2k+2\right)}{4}$$

B. 1; 1²+2²+3³+k ²= $$\frac{\text{(k+k) (k+1)}}{2};\ \frac{\left(k+k\right)\left(k+1\right)\left(2k+1\right)}{3}$$

C. 1; 1²+2²+3³+k ²= $$\frac{\text{k(k+1) (2k+2)}}{6};\ \frac{\left(k+k\right)\left(k+1\right)\left(2k+3\right)}{3}$$

E. . 1; 1²+2²+3³+k ²= $$\frac{\text{k(k+1) (k+1)}}{-6};\ \frac{\left(k+1\right)\left(k+1\right)\left(2k+3\right)}{-6}$$

Buktikanlah bahwa jumlah n bilangan asli pertama adalah 1+2+3+...+n = $$\frac{1}{2}$$ 𝑛[𝑛 + 1]

B. 1; 1+2+3+...+k+ $$\frac{1}{2}$$ 𝑘[𝑘 + 1]; (𝑘 +1)( $$\frac{1}{2}$$ 𝑘 + 1)

A. 2; 1+2+3+...+k+ $$\frac{1}{2}$$ 𝑘[𝑘 + 1]; (𝑘 + 1)( $$\frac{1}{2}$$ 𝑘 + 2

C. -1; 1+2+3+...+k+ $$\frac{1}{4^2}$$ 𝑘[𝑘 + 1]; (𝑘 + 1)( $$\frac{1}{2^2}^{ }$$ 𝑘 -1

D. 1; 1+2+3+...+k+ $$\frac{1}{2}$$ [𝑘 - 1]; (𝑘 - 1)( $$\frac{1}{2}$$ + 1

E. 1; 1+2+3+...+k+ $$\frac{1}{2}$$ 𝑘[2k + 1] ; (3𝑘 + 2)( $$\frac{1}{2}$$ 𝑘 + 1

ULANGAN INDUKSI MATEMATIKA

Quiz

•

Mathematics

•

11th Grade

•

Practice Problem

•

Hard

Nureskya Almafira

Used 8+ times

FREE Resource

10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

5 mins • 10 pts

Dengan induksi matematika buktikanlah rumus 2+4+6+...+2n=n²+n (untuk: n=1; n= k ; n= k+1)

A, 1²+1 ; 2+4+6+... 2(k)= k+k ; (k+1) (k+1)²

B. 1²+1 ; 2+4+6+... 2(k)= k²+k ; (k+1)² (k+1)

C. 1+1 ; 2+4+6+... 2(k)= k²+k ; 2(k+1)

D. 1²+1 ; 2+4+6+... 2(k)= k+k²; (k+1) (k+1)

E. 1²+1 ; 2+4+6+... 2(k)= k²+k ; (k+1) (k+1)

2.

MULTIPLE CHOICE QUESTION

5 mins • 10 pts

Diketahui 1+7+13+19+...+(6n-5)=3n²-2n berlaku untuk semua bilangan asli n. Berdasarkan prinsip induksi matematika: untuk n = 1, ruas kiri = 1, dan ruas kanan = 3(1)² - 2(1) = 3-2 = 1. Jadi, P(1) benar. Andaikan P(k) benar, maka P(k + 1) = ...

A. 1+7+13+19+...+(6k-5)=3k²-2k

B. 1+7+13+19+...+(6k-1)=3k²-4k-1

C. 1+7+13+19+...+(6k-1)=3k²+4k

D. 1+7+13+19+...+(6k+1)=3k²-4k

E. 1+7+13+19+...+(6k+1)=3k²+4k+1

3.

MULTIPLE CHOICE QUESTION

5 mins • 10 pts

Tunjukkan bahwa P(n)=n³ +5n habis di bagi 3 untuk setiap n bilangan asli (untuk: n=1 ;n= k; n=k+1)

A. 6 ; k³+5k; 5(k²+k+4)

B. 3 ; k³+5k² ; 3(k+k+1)

C. ; k³+5k ; 3(k²+k+2)

D. 6 ; k³ +5k ; 6( k²+k²+2)

E. 9; k³+ 5 ; 3(k+k²+2)

4.

MULTIPLE CHOICE QUESTION

5 mins • 10 pts

Dengan induksi matematika buktikan bahwa 3+7+11+ (4n-1)= 2n²+ n Dimana n adalah bilangan asli ( untuk n=1, n=k, n= k+1 )

A. 3 ; 3+7+11+ (4k-1) = 2k²+k ; 2k² +5k+ 3

B. 5 ; 3 +7+11+ (4k+1) = 2k²+k ; 3k²+10k +3²

C. 3 ; 3+7+11+ (4k-1) = 2k²+k ; 2k²+5k+6

D. 4 ; 3+7+11+ (-4k+1) = 2k²+k² ; 6k² +4k+ 5

E. 5 ; 3+7+11+ (4k²-1) = 2k²+k ; 4k² +5k+ 6

5.

MULTIPLE CHOICE QUESTION

5 mins • 10 pts

6.

MULTIPLE CHOICE QUESTION

5 mins • 10 pts

Dengan Induksi matematika buktikanlah bahwa 5+7+9+...+(2n+3) = n²+4n

A. 5; 5+7+9+...+(2k+3) =k²+4k; k² + 6k + 5

B. -5; 5+7+9+...+(2k+3) =k²+4k; k² + 6k + 5

C. 5; 5+7+9+...+(2k²+3) =k²+4k; k² + 6k + 5

D. 5; 5+7+9+...+(2k+3) =k²+4k; k² + 6k² + 5

E, 5; 5+7+9+...+(2k+3) =k²+4k; k² + 6k + 5²

7.

MULTIPLE CHOICE QUESTION

5 mins • 10 pts

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