Q. 8 D4.7( 3 Votes )

# Find the sum of t

Generalized term be (2r - 1)2=4r2 + 1 - 4r

1st term = 4(1)2 + 1 - 4(1)

2nd term =4(2)2 + 1 - 4(2)

And so on

nth term= 4n2 + 1 - 4n

Summation=1st term + 2nd term + …….. + nth term

= 4(1)2 + 1 - 4(1) + 4(2)2 + 1 - 4(2) ………4n2 + 1 - 4n ……(1)

We know ,

Thus

From (1) we have

Summation =

We know by property that:

∑axn + bxn - 1 + cxn - 2…….d0=a∑xn + b∑xn - 1 + c∑xn - 2…….. + d0∑1

Thus

(2)

We know

Thus substituting above values in (2)

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