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If $|{a_k}| < 1,{\lambda _k} \ge 0$ for $k = 1,\,2,....n$ and ${\lambda _1} + {\lambda _2} + ... + {\lambda _n} = 1,$ then the value of $|{\lambda _1}{a_1} + {\lambda _2}{a_2} + ....{\lambda _n}{a_n}|$ is
Equal to one
Greater than one
Zero
Less than one
Solution
(d) We have $|{\lambda _1}{a_1} + {\lambda _2}{a_2} + ….. + {\lambda _n}{a_n}|$
$ \le |{\lambda _1}{a_1}| + |{\lambda _2}{a_2}| + ….. + |{\lambda _n}{a_n}|$
$ = |{\lambda _1}||{a_1}| + ….. + |{\lambda _n}||{a_n}|$
$ = {\lambda _1}|{a_1}| + ….. + {\lambda _n}|{a_n}|\,$[ each ${\lambda _k}_. \ge 0$]
$ < {\lambda _1} + ….. + {\lambda _n}$
[ $|{a_k}| < $ $1$ and so ${\lambda _k}|{a_k}| < {\lambda _k}$for all $k = 1,2,….n$]
Hence $|{\lambda _1}{a_1} + {\lambda _2}{a_2} + ….. + {\lambda _n}{a_n}| < 1$.
Thus $|{\lambda _1}{a_1} + ….. + {\lambda _n}{a_n}| < 1$.
