Q.Evaluate $(i^{131}+i^{49})$

The imaginary unit $i$ has the defining property $i^2=-1$, and its higher powers cycle through just four values: $i^1=i$, $i^2=-1$, $i^3=i\cdot i^2=-i$, and $i^4=i^2\cdot i^2=1$, after which the pattern repeats — $i^5=i$, $i^6=-1$, and so on. This four-value cycle is the key to evaluating $i^n$ for any integer $n$, however large: divide $n$ by $4$ to get $n=4m+r$ with remainder $0\le r<4$, and then $i^n=i^{4m}\cdot i^r=(i^4)^m\cdot i^r=1^m\cdot i^r=i^r$ — the whole power collapses to whichever of $1,i,-1,-i$ corresponds to the remainder. In compact form: $i^{4k}=1$, $i^{4k+1}=i$, $i^{4k+2}=-1$, $i^{4k+3}=-i$ for any integer $k$ (this works for negative exponents too, using $i^{-1}=\dfrac1i …