ak=cos(kπ7)+isin(kπ7)=ei(kπ7)
So,
ak+1−ak=ei((k+1)π7)−ei(kπ7)=ei(kπ7)(eiπ7−1)=ei(kπ7)eiπ14(eiπ14−e−iπ14)=ei((2k+1)π7)×[2isin(π14)]⇒|ak+1−ak|=2sin(π14)
Replacing k with 4k−2 in the above expression,
|a4k−1−a4k−2|=2sin(π14)
Hence the value of the expression,
12∑k=1|a(k+1)−ak|3∑k=1|a(4k−1)−a(4k−2)|=246=4
Alternate Solution
ak=cos(kπ7)+isin(kπ7)
|ak+1−ak|=∣∣
∣∣(cos(k+1)π7−coskπ7)+i(sin(k+1)π7−sinkπ7)∣∣
∣∣
=
⎷2−2(cos(k+1)π7coskπ7+sin(k+1)π7sinkπ7)
⇒|ak+1−ak|=√2−2cosπ7
=√2(1−cosπ7)
=2sinπ14
Replace k with 4k−2 in above equation
⇒|a4k−2−a4k−2|=2sinπ14
Hence the value of the expression,
12∑k=1|a(k+1)−ak|3∑k=1|a(4k−1)−a(4k−2)|=246=4