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## Answers

=(1+3+5+7....n+1 terms) - (2+4+6+....n terms)

=[(1+2+3+4+5+....2n+1 terms) - (2+4+6+....n terms)] - (2+4+6+....n terms)

= (1+2+3+4+5+....2n+1 terms) - 2x(2+4+6+....n terms)

= (1+2+3+4+5+....2n+1 terms) - 2x2x(1+2+3+4+5+....n terms)

= [(2n+1)(2n+2) / 2 ] - [4 x n(n+1) / 2]

= (2n+1)(n+1) - 2n(n+1) = (n+1)[(2n+1) - 2n] = (n+1)

in terms of (2n+1), (n+1) = [(2n+1)/2] - 1/2

1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + 9 - 10 ..............

Consider the terms in pairs

(1 - 2) + (3 - 4) + (5 - 6) + (7 - 8) + (9 - 10) ..............

-1 + -1 + -1 + -1 + -1 .........

Sum of these terms for (2n+1) terms is –(2n+1) = –2n – 1