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NCERT Exemplar · Q2

Q.Evaluate ∑n=113(in+in+1)\displaystyle\sum_{n=1}^{13}\left(i^n+i^{n+1}\right), where n∈Nn\in\mathbf{N}.

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The sum simplifies by pairing terms: in+in+1=in(1+i)i^n + i^{n+1} = i^n(1+i). Over 13 terms, the pattern repeats every 4, and the total evaluates to −1+i-1 + i.

Why this approach works

When you see a sum of powers of ii, the first instinct should be to look for periodicity. The imaginary unit ii has a clean cycle: i1=ii^1 = i, i2=−1i^2 = -1, i3=−ii^3 = -i, i4=1i^4 = 1, and then it repeats every 4. That means any sum over consecutive powers can be grouped into blocks of 4, and each block often cancels or gives a constant.

But here we have a twist: each term is actually a pair — in+in+1i^n + i^{n+1}. That’s not just two separate powers; it’s a sum of two consecutive powers. If you factor ini^n out of that pair, you get in(1+i)i^n(1 + i). That’s a huge simplification: now the whole sum becomes (1+i)(1+i) times the sum of ini^n from n=1n=1 to 1313.

So the problem reduces to: what is S=∑n=113inS = \sum_{n=1}^{13} i^n? Then multiply by (1+i)(1+i).

Step-by-step solution

1. Factor the pair

For any nn,

in+in+1=in(1+i).i^n + i^{n+1} = i^n(1 + i).

This is true because in+1=in⋅ii^{n+1} = i^n \cdot i. So the entire sum becomes

∑n=113(in+in+1)=(1+i)∑n=113in.\sum_{n=1}^{13} (i^n + i^{n+1}) = (1+i) \sum_{n=1}^{13} i^n.

2. Sum the powers of ii from n=1n=1 to 1313

The powers of ii repeat every 4:

i1=i,i2=−1,i3=−i,i4=1,i^1 = i,\quad i^2 = -1,\quad i^3 = -i,\quad i^4 = 1,

and then i5=ii^5 = i, i6=−1i^6 = -1, etc.

So the sum of one full cycle of 4 terms is

i+(−1)+(−i)+1=0.i + (-1) + (-i) + 1 = 0.

That’s a key observation: every block of 4 consecutive powers of ii sums to zero.

3. Break 13 terms into cycles

From n=1n=1 to 1313, we have:

  • Three full cycles: n=1n=1 to 44, 55 to 88, 99 to 1212 — each sums to 00.
  • That leaves one leftover term: n=13n=13.

Since 13÷413 \div 4 gives remainder 11, i13=i1=ii^{13} = i^{1} = i (because 13=4⋅3+113 = 4\cdot3 + 1).

Therefore,

∑n=113in=0+0+0+i=i.\sum_{n=1}^{13} i^n = 0 + 0 + 0 + i = i.

Tip

A quick way: the sum of ini^n from n=1n=1 to NN is 00 if NN is a multiple of 4, ii if remainder 1, −1-1 if remainder 2, −i-i if remainder 3. Here N=13N=13 gives remainder 1, so the sum is ii.

4. Multiply by (1+i)(1+i)

Now we have

∑n=113(in+in+1)=(1+i)⋅i.\sum_{n=1}^{13} (i^n + i^{n+1}) = (1+i) \cdot i.

Compute:

(1+i)i=i+i2=i−1=−1+i.(1+i)i = i + i^2 = i - 1 = -1 + i.

Watch out

A common mistake is to forget that i2=−1i^2 = -1, not 11. Double-check: i⋅i=i2=−1i \cdot i = i^2 = -1, so 1⋅i+i⋅i=i+(−1)1 \cdot i + i \cdot i = i + (-1).

5. Final result

The sum is −1+i-1 + i, which in standard form is i−1i - 1.

✓Final answer

The value of the sum is −1+i\boxed{-1 + i}.

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