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  • Find the sum to n terms of the series in Exercises 8 to 10 whose nth terms is given by 2n minus 1 square

10.   Find the sum to n terms of the series in Exercises 8 to 10 whose nth terms is given by  ( 2n-1) ^2

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nth terms is given by  ( 2n-1) ^2.

a_n=( 2n-1) ^2=4n^2+1-4n

S_n=\sum _{k=1}^{n} a_k=\sum _{k=1}^{n} (2k-1)^2

                        =4\sum _{k=1}^{n} k^2-4\sum _{k=1}^{n} k+\sum _{k=1}^{n} 1

                     =\frac{4.n(n+1)(2n+1)}{6}-\frac{4.n(n+1)}{2}+n

                   =\frac{2.n(n+1)(2n+1)}{3}-2.n(n+1)+n

                    =n[\frac{2(n+1)(2n+1)}{3}-2(n+1)+1]

                   =n(\frac{4n^2+6n+2-6n-6+3}{3})

                     =n(\frac{4n^2-1}{3})

                     =n(\frac{(2n+1)(2n-1)}{3}) 

Posted by

seema garhwal

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