CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

If (1+x)n=P0+P1x+P2x2+...+Pnxn,

then p0p2+p4p6+...p1p3+p5p7+...=?

A
itannπ4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
tannπ4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
cotnπ4
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
icotnπ4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C cotnπ4
Let x=i
Hence, (1+i)n=P0+P1i+P2i2+P3i3...+Pnin
(1+i)n=P0+P1iP2P3i+...
(1+i)n=(P0P2+P4P6+...)+i(P1P3+P5...)
(2)n.einπ4=(P0P2+P4+...)+i(P1P3+P5...)
Comparing real and imaginary parts,
P0P2+P1...=(2)2cosnπ4------(1)
and P1P3+P5....=(2)2sinnπ4----------(2)
So, (1)(2)= required expression =cot(nπ4)

flag
Suggest Corrections
thumbs-up
0
mid-banner-image
mid-banner-image
similar_icon
Related Videos
thumbnail
lock
De-Moivre's Theorem
MATHEMATICS
Watch in App