CBSE Class 11-science Answered

Dear Student,

Let us examine the 1^st terms of each row.

1^st row........1

2^nd row.......3 ..........(1+2.1)

3^rd row.......7............(1+2.1+2.2)

4^th row......13..........(1+2.1+2.2+2.3)

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n^th row..................(1+2.1+2.2+2.3+........+2.(n-1))

Hence. 1^st element of n^th 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:

1^st row........1

2^nd row.......5 ..........(1+2.2)

3^rd row.......11............(1+2.2+2.3)

4^th row......19..........(1+2.2+2.3+2.4)

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n^th 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 n^th 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.