Iklan

Pertanyaan

Buktikanlah. c. 1 ⋅ 2 0 + 2 ⋅ 2 1 + 3 ⋅ 2 2 + 4 ⋅ 2 3 + .... + n ⋅ 2 n − 1 = ( n − 1 ) ⋅ 2 n + 1

Buktikanlah.

c.  

Ikuti Tryout SNBT & Menangkan E-Wallet 100rb

Habis dalam

01

:

15

:

15

:

00

Klaim

Iklan

A. Acfreelance

Master Teacher

Mahasiswa/Alumni UIN Walisongo Semarang

Jawaban terverifikasi

Pembahasan

Membuktikan dengan menggunakan induksi matematika dimana Untuk n =1 Untuk n =k diasumsikan terbukti Untuk n = k+1 maka Maka akan dibuktikan Jadi terbukti karena hasil sisi kanan dan kiri sama

Membuktikan dengan menggunakan induksi matematika dimana

Untuk n =1

table attributes columnalign right center left columnspacing 0px end attributes row cell 1 times 2 to the power of 0 plus 2 times 2 to the power of 1 plus 3 times 2 squared plus 4 times 2 cubed plus.... plus straight n times 2 to the power of straight n minus 1 end exponent end cell equals cell left parenthesis straight n minus 1 right parenthesis times 2 to the power of straight n plus 1 end cell row cell 1 times 2 to the power of 0 plus 2 times 2 to the power of 1 plus 3 times 2 squared plus 4 times 2 cubed plus.... plus 1 times 2 to the power of 1 minus 1 end exponent end cell equals cell left parenthesis 1 minus 1 right parenthesis times 2 to the power of 1 plus 1 end cell row cell 1.2 to the power of 0 end cell equals cell 0.2 plus 1 end cell row 1 equals cell 1 rightwards arrow terbukti end cell end table

Untuk n =k diasumsikan terbukti 

table attributes columnalign right center left columnspacing 0px end attributes row cell 1 times 2 to the power of 0 plus 2 times 2 to the power of 1 plus 3 times 2 squared plus 4 times 2 cubed plus.... plus straight n times 2 to the power of straight n minus 1 end exponent end cell equals cell left parenthesis straight n minus 1 right parenthesis times 2 to the power of straight n plus 1 end cell row cell 1 times 2 to the power of 0 plus 2 times 2 to the power of 1 plus 3 times 2 squared plus 4 times 2 cubed plus.... plus straight k times 2 to the power of straight k minus 1 end exponent end cell equals cell left parenthesis straight k minus 1 right parenthesis times 2 to the power of straight k plus 1 rightwards arrow terbukti end cell end table

Untuk n = k+1 maka

table attributes columnalign right center left columnspacing 0px end attributes row cell 1 times 2 to the power of 0 plus 2 times 2 to the power of 1 plus 3 times 2 squared plus 4 times 2 cubed plus.... plus straight n times 2 to the power of straight n minus 1 end exponent end cell equals cell left parenthesis straight n minus 1 right parenthesis times 2 to the power of straight n plus 1 end cell row cell 1 times 2 to the power of 0 plus 2 times 2 to the power of 1 plus 3 times 2 squared plus 4 times 2 cubed plus.... plus open parentheses straight k plus 1 close parentheses times 2 to the power of straight k minus 1 plus 1 end exponent end cell equals cell left parenthesis straight k plus 1 minus 1 right parenthesis times 2 to the power of straight k plus 1 end exponent plus 1 end cell row cell 1 times 2 to the power of 0 plus 2 times 2 to the power of 1 plus 3 times 2 squared plus 4 times 2 cubed plus.... plus open parentheses straight k plus 1 close parentheses times 2 to the power of straight k end cell equals cell left parenthesis straight k right parenthesis times 2 to the power of straight k plus 1 end exponent plus 1 end cell row cell 2 straight k to the power of straight k plus 2 to the power of straight k end cell equals cell 2 straight k to the power of straight k plus 1 end exponent plus 1 rightwards arrow terbukti end cell end table

Maka akan dibuktikan

table attributes columnalign right center left columnspacing 0px end attributes row cell 1 times 2 to the power of 0 plus 2 times 2 to the power of 1 plus 3 times 2 squared plus 4 times 2 cubed plus.... plus straight n times 2 to the power of straight n minus 1 end exponent end cell equals cell left parenthesis straight n minus 1 right parenthesis times 2 to the power of straight n plus 1 end cell row cell 1 times 2 to the power of 0 plus 2 times 2 to the power of 1 plus 3 times 2 squared plus 4 times 2 cubed plus.... plus k.2 to the power of k plus open parentheses straight k plus 1 close parentheses times 2 to the power of straight k minus 1 plus 1 end exponent end cell equals cell left parenthesis straight k plus 1 minus 1 right parenthesis times 2 to the power of straight k plus 1 end exponent plus 1 end cell row cell 1 times 2 to the power of 0 plus 2 times 2 to the power of 1 plus 3 times 2 squared plus 4 times 2 cubed plus.... plus 1 left parenthesis straight k minus 1 right parenthesis.2 to the power of straight k plus 2 straight k to the power of straight k plus 2 to the power of straight k end cell equals cell left parenthesis straight k right parenthesis times 2 to the power of straight k plus 1 end exponent plus 1 end cell row cell 1 plus 2 straight k to the power of straight k minus 2 to the power of straight k plus 2 straight k to the power of straight k plus 2 to the power of straight k end cell equals cell 2 straight k to the power of straight k plus 1 end exponent plus 1 end cell row cell 1 plus 4 straight k to the power of straight k end cell equals cell 2 straight k to the power of straight k plus 1 end exponent plus 1 end cell row cell 1 plus 2 straight k.2 to the power of straight k end cell equals cell 2 straight k to the power of straight k plus 1 end exponent plus 1 end cell row cell 1 plus straight k.2 to the power of 1.2 to the power of straight k end cell equals cell 2 straight k to the power of straight k plus 1 end exponent plus 1 end cell row cell 2 straight k to the power of straight k plus 1 end exponent plus 1 end cell equals cell 2 straight k to the power of straight k plus 1 end exponent plus 1 rightwards arrow terbukti end cell end table

Jadi terbukti 1 times 2 to the power of 0 plus 2 times 2 to the power of 1 plus 3 times 2 squared plus 4 times 2 cubed plus.... plus straight n times 2 to the power of straight n minus 1 end exponent equals left parenthesis straight n minus 1 right parenthesis times 2 to the power of straight n plus 1 karena hasil sisi kanan dan kiri sama

 

Perdalam pemahamanmu bersama Master Teacher
di sesi Live Teaching, GRATIS!

1

Eka Apriliya

Jawaban tidak sesuai

Iklan

Pertanyaan serupa

Gunakan prinsip induksi matematika untuk membuktikan setiap notasi sigma berikut. a. k = 1 ∑ n ​ k 2 + k 1 ​ = n + 1 n ​

1

1.0

Jawaban terverifikasi

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Roboguru

Coba GRATIS Aplikasi Ruangguru

Download di Google PlayDownload di AppstoreDownload di App Gallery

Produk Ruangguru

Hubungi Kami

Ruangguru WhatsApp

+62 815-7441-0000

Email info@ruangguru.com

[email protected]

Contact 02130930000

02130930000

Ikuti Kami

©2026 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia