Find the sum of the first n odd integers.?

Question

Find the sum of the first n odd integers.
1 + 3 + 5 + … + 57

a) 435
b)406
c)1653
d)841

10pts for correct answer

Answer 1

These are the first 58/2 = 29 odd integers, so the sum is the square of 29, which is 841.

Answer 2

The answer is n^2 (n-squared)

n=1 1 = 1
n=2 1+3 = 4
n=3 1+3+5 = 9
n=4 1+3+5+7 = 16
n= 5 1+3+5+7+9 = 25

etc.

Answer 3

in case you initiate including numbers jointly, you will hit upon a million =a million a million+3 = 4 a million+3+5 = 9 a million+3+5+7 = sixteen The sum of the 1st n numbers is n^2. So what n is 201? a million is the 1st term, 3 is the 2nd term, 5 is the third term. in case you're taking the term, upload a million and divide via 2, then you particularly get the term selection. I.e. (5+a million)/2 = 3, so 5 is the third term. (201 +a million)/a million = one 0 one; one 0 one^2 = 10201.

Answer 4

tn = a + (n-1) d
57 = 1 + (n-1) 2
57 = 1 + (2n-2)
57 = 2n - 1
58 = 2n
29 = n

Sn = n/2 [2a+(n-1)d]
S(29) = 14.5 [2(1) + 28 (2)]
S(29) = 14.5 [58]
S(29) = 841

The answer is D. 841.

Answer 5

d) 841

Arithmetic progression
1, 3, 5..., 57
a=1, d=2

Tn = a+(n-1)d
57 = 1+(n-1)(2)
57-1 = (n-1)(2)
56/2 = n-1
28+1 = n
n = 29

Sn = (n/2)[2a+(n-1)d]
S29 = (29/2)[2(1)+(29-1)(2)]
S29 = (29/2)[2+56]
S29 = (29/2)(58)
S29 = 841

Answer 6

1 = 2*1 - 1
57 = 2*n - 1
n = 58/2 = 29
29
∑(2n - 1) =
1
. 29
(2∑n) - 29 =
. 1
29*30 - 29 =
29^2 = 841

In general,
n
∑(2i - 1) = n^2
i=1

Answer 7

sum of an arithmetic sequence
n/2(first term+last term
29/2(1+57
841

Answer 8

The formula is

sum = (first_term + last_term) * number_of_terms/2

Answer 9

None of the above.