Question

श्रेणी {1²-2²+3²-4²+ ...} 2n पदों का योग कया होगा?

Wrong answer not allow!

Genius mind users please help me.​

Answer 1

Answer:

Give a like

Step-by-step explanation:

1² - 2² + 3² - 4²

1 - 4 + 9 - 16

10 - 20

-10

Hope it will help you

Answer 2

Answer:

I am solving this problem directly

Step-by-step explanation:

प्रश्न :—

श्रेणी {1²-2²+3²-4²+ ...} 2n पदों का योग कया होगा?

ज्ञात करना है:—

पदों का योग कया होगा?

हल:—

1^2 - 2^2 + 3^2 - 4^2 + . . . . - (2n)^2

= [1^2 + 2^2 + 3^2 + 4^2 + . . . . + (2n)^2]

= 2[2^2 + 4^2 + 6^2 + . . . . + (2n)^2]

= 2[1^2 + 2^2 + 3^2 + 4^2 + . . . . + (2n)^2] - 2 × 2^2 [ 1^2 + 2^2 + 3^2 + . . . . + n^2]

= 2n(2n + 1) (2 × 2n + 1) /6 - 8 × n(n + 1)(2n + 1) /6

= 2n(2n + 1) (-3) /6

= - n(2n + 1) Ans.