CBSE Class 11-science Answered
Dear Student,
Let us examine the 1st terms of each row.
1st row........1
2nd row.......3 ..........(1+2.1)
3rd row.......7............(1+2.1+2.2)
4th row......13..........(1+2.1+2.2+2.3)
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nth row..................(1+2.1+2.2+2.3+........+2.(n-1))
Hence. 1st element of nth row is:
(1+2.(1+2+3+......+n)-2n)
=>(1+2.n(n+1)/2-2n)
=>(n2-n+1).
Similarly, analysing the last terms of each row:
1st row........1
2nd row.......5 ..........(1+2.2)
3rd row.......11............(1+2.2+2.3)
4th row......19..........(1+2.2+2.3+2.4)
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nth row.................(1+2.2+2.3+2.4+..........+2.n)
=>(1+2.(1+2+3+4+.....+n)-2)
=>(1+2.n.(n+1)/2-2)
=>The last term of nth row is (n2+n-1).
By inspection, for n= 45, first tem is1981, and last term is 2069 (within which 2009 is lying).
Regards Topperlearning.