Question

the sum of 1st 6 consecutive odd number.​

Answer 1

thirty six

it's 36

1+3+5+7+9+11=36

Answer 2

Answer:

First of all, I presume that you are referring to the first 6 positive odd numbers (of course 0 is even, so we start at 1). If we know (say) these numbers and since the numbers we need to sum is relatively small, then we can simply compute this sum by brute force, that is∑1+3+5+7+9+11=36=6⋅6=62. But (again) I assume that's not what you looking for. Suppose you know that set of positive odd numbers is given by {2k+1|k∈N}, where N=0,1,2,3,⋯. Let 2k+1 denotes the kth positive odd number, then 2(k+1)+1=2k+3 is the (k+1)thpositive odd number and Tk+1−Tk=2 and call that d

, to denote the common difference between (any) consecutive odd numbers (positive or negative for that matter). Now notice the following of sequence of positive odd numbers, i.e.

T0=2⋅0+1=1,T1=1+2⋅1=3,T2=1+2⋅2=5,⋯

Next, we shall consider the sequence of the kth

partial sums,call it Sk (say), k=1,2,3,…

i.e.

S1=T0=1=12,S2=T0+T1=1+3=4=22,S3=S2+T2=4+5=9=32,⋯

So we conjecture that a (nice) simple closed form formula for the sum of first k

positive odd numbers is given by Sk=k2,∀k=1,2,3,…

, which formula can be proven using mathematical induction.

⟹S6=62=6⋅6=36.

Aside:

To avoid the (possible) confusion between the positive odd numbers indexed from 0

,we can use instead the alternative formula {2k−1|k∈N}, where N=,1,2,3,⋯, to get the same results.