Question

Find the sum 1 x 3 x 5 + 3 x 5 x 7 + 5 x 7 x 9 + ... + (2n – 1) (2n + 1) (2n + 3) 

Answer 1

1 x 3 x 5 + 3 x 5 x 7 + 5 x 7 x 9 + ... + (2n – 1) (2n + 1) (2n + 3) 
Now, 1, 3, 5, 7, … are in A.P. with a = 1 and d = 2.
∴ r^th term = 1 + (r – 1)2 = 2r – 1
3, 5, 7, 9, … are in A.P. with a = 3 and d = 2
∴ r^th term = 3 + (r – 1)2 = 2r + 1
and 5, 7, 9, 11, … are in A.P. with a = 5 and d = 2.
∴ r^th term = 5 +(r – 1)2 = 2r + 3
∴ 1 x 3 x 5 + 3 x 5 x 7 + 5 x 7 x 9 + ... upto n terms

= \[\displaystyle\sum_{r=1}^{n}(2r - 1)(2r + 1)(2r + 3)\]

= \[\displaystyle\sum_{r=1}^{n}(4r^2 - 1)(2r + 3)\]

= \[\displaystyle\sum_{r=1}^{n} (8r^3 + 12r^2 - 2r - 3)\]

= 8\[\displaystyle\sum_{r=1}^{n} r^3 + 12\displaystyle\sum_{r=1}^{n}r^2 - 2\displaystyle\sum_{r=1}^{n} r - 3\displaystyle\sum_{r=1}^{n} 1\]

= `8{("n"("n" + 1))/2}^2 + 12{("n"("n" + 1)(2"n" + 1))/6} - 2{("n"("n" + 1))/2} - 3"n"`

= 2n²(n + 1)² + 2n(n + 1)(2n + 1) – n(n + 1) – 3n
= n(n + 1)[2n(n + 1) + 4n + 2 – 1] – 3n
= n(n + 1)(2n² + 6n + 1) – 3n
= n(2n³ + 8n² + 7n + 1 – 3)
= n(2n³ + 8n² + 7n – 2).