Sum of four consecutive powers of i(iota) is zero.
i.e.,`i^(n)+i^(n+1)+i^(n+2)+i^(n+3)=0,forall n in I.`
If `sum_(r=-2)^(95)i^(r)+sum_(r=0)^(50)i^(r!)=a+ib, " where " i=sqrt(-1)`, the unit digit of `a^(2011)+b^(2012)`, is

Options

A. (a)2
B. (b)3
C. (c)5
D. (d)6

Answer

To solve the problem step by step, we need to evaluate the two sums involving powers of \(i\) and factorials, and then find the unit digit of \(a^{2011} + b^{2012}\).

### Step 1: Evaluate the first sum
We start with the first sum:
\[ S_1 = \sum_{r=-2}^{95} i^r \]
We can break this down into groups of four, since we know that the sum of four consecutive powers of \(i\) is zero:
\[ i^0 + i^1 + i^2 + i^3 = 0 \]
Thus, we can group the terms as follows:
- From \(r = -2\) to \(r = 95\), we have \(98\) terms total.
- The first four terms are \(i^{-2}, i^{-1}, i^0, i^1\).
- The remaining terms can be grouped into sets of four.

Calculating the first four terms:
\[ i^{-2} = \frac{1}{i^2} = \frac{1}{-1} = -1, \quad i^{-1} = -i, \quad i^0 = 1, \quad i^1 = i \]
Thus,
\[ i^{-2} + i^{-1} + i^0 + i^1 = -1 - i + 1 + i = 0 \]
Now, we consider the remaining terms from \(i^2\) to \(i^{95}\). Since \(i^4 = 1\), we can group the terms in sets of four:
\[ (i^2 + i^3 + i^4 + i^5) + (i^6 + i^7 + i^8 + i^9) + \ldots + (i^{92} + i^{93} + i^{94} + i^{95}) = 0 \]
Each complete group of four sums to zero. The total number of terms from \(i^2\) to \(i^{95}\) is \(94\), which gives us \(23\) complete groups of four, plus the first four terms which sum to zero. Thus:
\[ S_1 = 0 \]

### Step 2: Evaluate the second sum
Next, we evaluate the second sum:
\[ S_2 = \sum_{r=0}^{50} i^{r!} \]
The factorial grows quickly, and we only need to consider \(r!\) modulo \(4\) because the powers of \(i\) repeat every four terms.

Calculating \(r!\) for \(r = 0\) to \(3\):
- \(0! = 1 \Rightarrow i^{0!} = i^1 = i\)
- \(1! = 1 \Rightarrow i^{1!} = i^1 = i\)
- \(2! = 2 \Rightarrow i^{2!} = i^2 = -1\)
- \(3! = 6 \Rightarrow i^{3!} = i^6 = i^2 = -1\)

For \(r \geq 4\), \(r!\) is divisible by \(4\), so:
- \(4! = 24 \Rightarrow i^{4!} = i^0 = 1\)
- \(5! = 120 \Rightarrow i^{5!} = i^0 = 1\)
- Continuing this, all factorials for \(r \geq 4\) will contribute \(1\).

Thus, for \(r = 0, 1, 2, 3\):
\[ S_2 = i + i - 1 - 1 + 1 \cdot (50 - 3) = 2i - 2 + 48 = 46 + 2i \]

### Step 3: Combine results
Now, we combine \(S_1\) and \(S_2\):
\[ S = S_1 + S_2 = 0 + (46 + 2i) = 46 + 2i \]
Thus, we have \(a = 46\) and \(b = 2\).

### Step 4: Find the unit digit of \(a^{2011} + b^{2012}\)
Now we need to find the unit digits:
- The unit digit of \(46^{2011}\): The unit digit of \(6^n\) is always \(6\) for \(n \geq 1\).
- The unit digit of \(2^{2012}\): The unit digits of powers of \(2\) cycle every \(4\): \(2, 4, 8, 6\).
- \(2012 \mod 4 = 0\) gives a unit digit of \(6\).

Now we add the unit digits:
\[ 6 + 6 = 12 \]
The unit digit of \(12\) is \(2\).

### Final Answer
The unit digit of \(a^{2011} + b^{2012}\) is \(2\).