Question

Determine the sum of the first k odd positive integers for a number of values of k. What generalizations occur to you? Are your inferences correct for all positive integers k?

Answer 1

Answer:

Let be $n_i$ the i-th odd positive integer number, $\sum_{i=1}^{k}{n_i}=k^2$

Step-by-step explanation:

Let be $n_i$ the i-th odd positive integer number

$k=1\\n_1=1\\S_1=n_1=1\\\\k=2\\n_1=1, n_2=3\\S_2=n_1+n_2=1+3=4\\\\ k=3\\n_1=1, n_2=3, n_3=5\\S_3=n_1+n_2+n_3=1+3+5=9\\\\\\k=4\\n_1=1, n_2=3, n_3=5, n_4=7\\S_4=n_1+n_2+n_3+n_4=1+3+5+7=16$

It seems that the sum of the first k odd positives integers is $k ^ 2$

Note that,

$n_1=1=1+2\times 0\\n_2=3=1+2\times 1\\n_3=5=1+2\times 2\\n_4=7=1+2\times 3\\n_5=9=1+2\times 4\\.\\.\\.\\n_k=2k-1=1+2\times (k-1)$

thus,

$\sum_{i=1}^{k}{n_i}=\sum_{i=1}^{k}{(2i-1)}=2\sum_{i=1}^{k}i-\sum_{i=1}^{k}1\\\\\sum_{i=1}^{k}{n_i}=2\frac{k(k+1)}{2}-k=k^2$