Find the first four partial sums and the nth partial sum of the sequence an. [Hint: Use a property of logarithms to write the nth term as a difference.] A n = 3 log ( n/ n + 1 )

Question

s1=

s2=

s3=

s4=

sn=

Answer 1

nth partial sum, s_n, is the sum of the first n terms of the sequence

s_n = 3[(log1 - log2) + (log2 - log3) + (log3 - log4) + ... + (log(n-1) - logn) + (logn - log(n+1))] = -3log(n+1)

So, s₁ = -3log2

s₂ = -3log3

s₃ = -3log4

s₄ = -3log5

Answer 2

N = 3 log(N) - 3 log(N+1)

N+1 = 3 log(N+1) - 3 log(N+2)

N+2 = 3 log(N+2) - 3 log(N+3)

....

N+k = 3 log (N+k) - 3 log(N+k+1)

so the sum of the first K terms is 3 log(N) - 3log(N+k+1)