Let ak=( k) i+ j and bk= i-( k) j, where k= 2k-1 2n+1, for some n N, n>5. Then the value of k=1n | ak |2 k=1n | bk |2 is
Answer
3
Solution
Here | ak |2 = 2 k + 1 = 2 k and | bk |2 = 1 + 2 k = cosec2 k.
∵ 2 + cosec2 = 1 2 2 = 4 2 2 = 4 cosec2 2 and 2 k = k+1, we get
2 k = 4 cosec2 k+1 - cosec2 k
Let T = k=1ncosec2 k. Since n+1 = 2n 2n + 1 = - 1, cosec2 n+1 = cosec2 1, so
k=1ncosec2 k+1 = k=2n+1cosec2 k = T
k=1n 2 k = 4T - T = 3T
∴ Required ratio = 3TT = 3
Answer: 3