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Question : 90 of 160
Marks:
+1,
-0
Solution:
‌‌ Given, ‌‌(ex+e5x)=a0+a1x+a2x2+...‌⇒(e−2x+e2x)=a0+a1x+a2x2+...‌⇒‌‌2[1+‌+‌+...]‌=a0+a1x+a2x2+...‌⇒‌‌a1=a3=a5=...=0‌∴‌‌2a1+23a3+25a5+...=0‌
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