Question

1^2-2^2+3^2-4^2.....n
find d sum of n terms of Ap​

Answer 1

Answer:

-n(n+1)/2

Step-by-step explanation:

1^2-2^2+3^2-4^2.....

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

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

=(-3) + (-7) + (-11) + ..... + [-(2n-1)]

= -(3+7+....+2n+1) ........(1)

3,7, ....(2n+1) is an AP with a=3, d=4 and no of terms is n/2

So Sum = (n/2)/2[3+2n-1]

=n/4(2n+2)

=n(n+1)/2 ........(2)

From (1) and (2)

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

Answer 2

I can't understand your question