5 second problem

Algebra Level 3

What is 1 + 3 + 5 + 7 + 9 + 11 + ..... + (n*2) - 1, in terms of n?

Like this problem if you solved it in less than 5 seconds.

2^n (n*(n+1)) * 3/2 4n /sqrt{n} n^2 n^4 (n*(n+1)) / 2 (n+1)(n-1)

4 solutions

Ashwin Padaki
May 12, 2015

If you add both ends, you get 2n. Keep adding both ends and the sum will be 2n each time. You do this n/2 times. 2n*n/2 = n^2.

But $1$ is not part of the series defined by $a_{n}=n^{2}-1$ , unless $n$ can take irrational values.

Omkar Kulkarni - 6 years, 1 month ago

The series is $2n - 1$ , not $n^2 -1$ .

Calvin Lin Staff - 6 years, 1 month ago

Mohammad Khaza
Mar 2, 2018

$1=1^2$

$1+3=2^2$

$1+3+5=3^2$

$1+3+5+7=4^2$

.........................

so, $1+3+5+.....+n=n^2$

Rama Devi
May 13, 2015

The above question can be easily solved. The given series is nothing but the sum of n odd natural numbers.Therefore the sum of the series is given by n^2.

Tasnim Rawat
May 13, 2015

Sn=n/2(a+L) where a = 1 and L = 2n -1 Sn = n/2(1 + 2n -1) = n/2 (2n) = n^2

5 second problem

4 solutions

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