Question

plz do the answer quickly ​

Answer 1

Step-by-step explanation:

☢️PLEASE MARK AS BRAINLIEST☢️

Answer 2

Answer:

$\underline{\underline{i-1}}$

Step-by-step explanation:

For this, we must consider that the sum of any four consecutive powers of i is always zero.

So,

$\displaystyle\sum_{n=1}^{13}\Big(i^n+i^{n+1}\Big)\ =\ \sum_{n=1}^{13}i^n\Big(1+i\Big)\ =\ \Big(1+i\Big)\sum_{n=1}^{13}i^n\\\\\\=\ (1+i)\left(\sum_{n=1}^{4}i^n+\sum_{n=5}^{8}i^n+\sum_{n=9}^{12}i^n+i^{13}\right)$

According to the concept, the three sums in the bracket become 0, so we get,

$(1+i)i^{13}$

Since  $i^{4n+k}=i^k,\quad\quad n,\ k\in\mathbb{Z},\quad\quad 0\leq k\leq 3,$

$(1+i)i^{13}\ =\ (1+i)i\ =\ \underline{\underline{i-1}}$