Question

Jumlah n suku pertama deret (k+1/k)+(k+3/k)+(k+5/k)+...=
A. n/2k(n+k)
B. n/2(n+k)
C. n/k(k+n)
D. n/2k(k-n)
E. 1/2k(k-n)

Jawan kalo yang tau . Pake caranya ya tnx

Answer 1

Verified answer

Nulisnya mungkin bisa lebih dirapihin lagi, biar lebih jelas.

((k+1)/k) + ((k+3)/k) + ((k+5)/k) + ...
= (1 + 1/k) + (1 + 3/k) + (1 + 5/k) + ...
= (1 + 1/k) + (1 + 1/k + 2/k) + (1 + 1/k + 2/k + 2/k) + ...

a = 1 + 1/k
b = 2/k

Sn = (n/2)(2a + (n - 1)b)
= (n/2)(2(1 + 1/k) + (n - 1)(2/k))
= (n/2)(2 + 2/k + 2n/k - 2/k)
= (n/2)(2 + 2n/k)
= n(1 + n/k)
= n((k + n)/k)
= (n/k)(k + n)