Find the sum of the following series. 1 + 5 + 9 + ... + 45?

Answer 1

a1=1
d=4
an=a1+(n-1)d
45=1+(n-1)4
n=12
Sn=n/2(a1+an)
S12=12/2(1+45)
S12=276

Answer 2

276

1 + 5 + 9 + 13 + 17 + 21 + 25 + 29 + 33 + 37 + 41 + 45 =276

Answer 3

1 + 5 + 9 + ... + 41 + 45
(1 + 45) + (5 + 41) + (9 + 37) + ...
This is 46 added together for every pair of numbers in your series.
An = A1 + d(n - 1)
45 = 1 + 4(n - 1)
44 = 4(n - 1)
11 = n - 1
n = 12. There are 12 terms, so there are 6 pairs.

S = 6(46) = 276.

Answer 4

That's an arithmetic series with a common difference of 4, and you have (45 - 1) / 4 + 1 = 12 terms, averaging (45 + 1) / 2= 23.
Total 12 * 23 = 276.

Answer 5

255 is the sum of 1 + 5+9+13+17+21+25+29+33+37+41+45.

Answer 6

1+5+9+13+17+21+25+29+33+37+41+45 = 276

Answer 7

x = 1 + 5 + 9 + ... + 37 + 41 + 45

x = 45 + 41 + 37 + ... + 3 + 2 + 1

2x = 46 + 46 + 46 + ... + 46 + 46 + 46

2x = 46*12 (there are 12 terms, i.e., (45 + 3)/4 = 12)

x = 46*12/2 = 46*6 = 276

Answer 8

This series can be written as:

[k=0, 11]∑(1+4k)

Breaking up the sum:

[k=0, 11]∑1 + 4[k=0, 11]∑k

Using the formulas for the sum of the first n numbers:

12 + 4(11)(12)/2
12 + 264
276

So the sum is in fact 276.

Answer 9

You have several groups of 46 (1+45, 5+41, 9+37, etc). count the number of groups and multiply by 46.
1+45,5+41,9+37,13+33,17+29 and 21+25. so there are 6 groups of 46. 46 times 6 =276

Answer 10

Let S=1+5+9+...+45
Notice that
S-12=0+4+8+...+44
(S-12)/4=0+1+2+...+11
(S-12)/4=11+10+9+...+0
(S-12)/4+(S-1)/4=(0+11)+(1+10)
+...+(11+0) <12 terms>
(S-12)/2=12*11=132
S-12=264
S=276

Answer 11

276. It appears adding 4 each time is the pattern in the question.