Question

Fund the sum of the series 1-2+3-4+5-6+....... to n terms​

Answer 1

1/2 is the answer hope it's helpful to you

Answer 2

Answer:

Step-by-step explanation:

1 - 2 + 3 - 4 + 5 - 6 + ... to n terms

Here if we observe we get,

⇒ ( 1 + 3 + 5 + 7 + ... ) + ( - 2 - 4 - 6 - 8 - ... )

⇒ ( 1 + 3 + 5 + 7 + ... ) - ( 2 + 4 + 6 + 8 + ... )

⇒ ( 1 + 3 + 5 + 7 + ... ) - ( 1 + 1 + 2 + 2 + 3 + 3 + 4 + 4 + ... )

[ Since 2 = 1 + 1 , 4 = 2 + 2 , 6 = 3 + 3, etc. ]

⇒ ( 1 + 3 + 5 + 7 + ... ) - 2 ( 1 + 2 + 3 + 4 + ... )

We know that,

• Sum of all odd numbers can be calculated by the formula: n²
• Sum of all Natural Numbers = [ n ( n + 1 ) / 2 ]

Substituting we get,

⇒ n² - 2  [ n ( n - 1 ) / 2 ]

⇒ n² - [ 2n ( n - 1 ) / 2 ]

⇒ n² - n ( n - 1 )           [ Since 2 and 2 gets cancelled, we get 'n' ]

⇒ n² - n² + n

⇒ n

This is the required answer.

Hope it is correct !