Question

Bantu jawaab please​

Answer 1

Jawab:

induksi

Penjelasan dengan langkah-langkah:

1/(1.3) + 1/(3.5) + ...+ 1/(2n-1) (2n+1) = n/(2n +1)

a(n) = 1/ { (2n- 1)(2n+1)}

a(k) = 1/ { (2n- 1)(2n+1)}

a(k+1) = 1/ {(2(k+1) -1)(2 (k+1) + 1)

a(k+1) = 1/{(2k+1)(2k+ 3)

.

p(n) = n/ (2n + 1)

p(k) = k/(2k + 1)

p(k+1) = (k+ 1) / (2(k+1) +1)

p(k+1) = (k+1)/(2k+ 3).

....

p(k+1)  = p(k) + a(k+1)

(k+1)/ (2k+3) =  k/(2k+1) +   1/ (2k +1)(2k+3)

(k+1)/ (2k+3) =  [ k (2k+3) + 1  ]/ (2k + 1)(2k+3)

(k+1)/ (2k+3) = [2k² + 3k + 1 ] / (2k + 1)(2k+ 3)

(k+1)/ (2k + 3)  = (2k +1)(k +1)/(2k+1)(2k+3)

(k+1) /  (2k+ 3) = (k+1) / (2k + 3)

....