Question

The value of $\sum\limits_{n=0}^{100}{{{i}^{n!}}}$ equals (where, $i=\sqrt{-1}$ ).

Answer 1

Hint: Expand the value of ${{i}^{n!}}$ from n = 0 to n = 100. Thus modify them by applying their factorial value. Now substitute the values of the power of iota and simplify it.

Complete step-by-step answer:
We have been given an expression for which we should find the value.
Given to us, $\sum\limits_{n=0}^{100}{{{i}^{n!}}}$.
First let us expand ${{i}^{n!}}$, where n = 0, 1, 2, 3,…….., 100 as it is given from n = 0 to n = 100.
Hence we can write it as,
$\sum\limits_{n=0}^{100}{{{i}^{n!}}}={{1}^{0!}}+{{i}^{1!}}+{{i}^{2!}}+{{i}^{3!}}+......+{{i}^{100!}}-(1)$
We know the basics of factorial,

& 0!=1 \\\ & 1!=1 \\\ & 2!=2\times 1=2 \\\ & 3!=3\times 2\times 1=6 \\\ & 4!=4\times 3\times 2\times 1=24 \\\ & 5!=5\times 4\times 3\times 2\times 1=120 \\\ \end{aligned}$$ And thus it continues like this. Hence, we can modify (1) on the above basis as, $$\sum\limits_{n=0}^{100}{{{i}^{n!}}}={{1}^{1}}+{{i}^{2}}+{{i}^{24}}+{{i}^{120}}+......+{{i}^{100!}}-(2)$$ We know that, $${{i}^{2}}=-1$$ and we have been given, $$i=\sqrt{-1}$$. Similarly, $${{i}^{4}}=1,{{i}^{6}}=-1$$ Thus we can substitute these values in (2). $$\begin{aligned} & \sum\limits_{n=0}^{100}{{{i}^{n!}}}=i+i+\left( -1 \right)+\left( -1 \right)+1+1+1+......+1 \\\ & \sum\limits_{n=0}^{100}{{{i}^{n!}}}=i+i-1-1+1+1+......+1 \\\ \end{aligned}$$ $$\sum\limits_{n=0}^{100}{{{i}^{n!}}}=2i-2+1+1+1+.....+1$$, here (+1) is repeated 97 times $$\begin{aligned} & \sum\limits_{n=0}^{100}{{{i}^{n!}}}=2i-2+97 \\\ & \sum\limits_{n=0}^{100}{{{i}^{n!}}}=2i+95 \\\ \end{aligned}$$ Hence, we got $$\sum\limits_{n=0}^{100}{{{i}^{n!}}}=2i+95$$. Thus we got the value of the given expression as $$\left( 2i+95 \right)$$. Note: To solve this question, you should know the concept of factorial as well as complex numbers. Expand the given expression from n = 0, 1, 2,…..,100. After this it will be easy to substitute factorial value. In $${{i}^{3!}},{{i}^{4!}},....{{i}^{100!}}$$ all their values will be (+1).