1.If Cr=nCrC_r = {}^nC_rCr=nCr, then C0+C4+C8+C12+…=C_0 + C_4 + C_8 + C_{12} + \ldots =C0+C4+C8+C12+…=a.2n/22[sinnπ4+2n/2−1]\frac{2^{n/2}}{2}[\sin\frac{n\pi}{4} + 2^{n/2-1}]22n/2[sin4nπ+2n/2−1]b.2n/2sinnπ42^{n/2}\sin\frac{n\pi}{4}2n/2sin4nπc.2n−1cosnπ42^{n-1}\cos\frac{n\pi}{4}2n−1cos4nπd.2n/22[cosnπ4+2n/2−1]\frac{2^{n/2}}{2}[\cos\frac{n\pi}{4} + 2^{n/2-1}]22n/2[cos4nπ+2n/2−1]Login to continueOnly logged in users canattempt or see the solution.
