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Question

Mathematics Question on Complex Numbers and Quadratic Equations

The value of ∑k=0n(ik+ik+1)\sum\limits^{n}_{k=0}\left(i^{k}+i^{k+1}\right), where i2=−1i^2 = -1, is equal to

A

i−ini-i^n

B

−i−in+1-i-i^{n+1}

C

i−in+1i-i^{n+1}

D

i−in+2i-i^{n+2}

Answer

i−in+2i-i^{n+2}

Explanation

Solution

∑k=0n(ik+ik+1)=∑k=0nik(i+1)\displaystyle\sum_{k=0}^{n}\left(i^{k}+i^{k+1}\right)=\displaystyle\sum_{k=0}^{n} i^{k}(i+1)
=(i+1)[1+i1+…+in]=(i+1)\left[1+i^{1}+\ldots+i^{n}\right]
=(i+1)[1(1−in+1)(1−i)]×1+i1+i=(i+1)\left[\frac{1\left(1-i^{n+1}\right)}{(1-i)}\right] \times \frac{1+i}{1+i}
=(1+i)2[1−in+1]12+1=(1−1+2i)(1−in+1)2=\frac{(1+i)^{2}\left[1-i^{n+1}\right]}{1^{2}+1}=\frac{(1-1+2 i)\left(1-i^{n+1}\right)}{2}
=2i2(1−in+1)=i−in+2=\frac{2 i}{2}\left(1-i^{n+1}\right)=i-i^{n+2}