Question

Find the sum 5 × 7 + 9 × 11 + 13 × 15 + ... upto n terms.

Answer 1

5 × 7 + 9 × 11 + 13 × 15 + ... upto n terms

Now, 5, 7, 9, 11, … are in A.P.

r^th term = 5 + (r – 1) (2) = 2r + 3

7, 9, 11, .… are in A.P.

r^th term = 7 + (r – 1) (2) = 2r + 5

∴ 5 × 7 + 7 × 9 + 9 × 11 × 13 + ... upto n terms

= \[\displaystyle\sum_{r=1}^{n} (2r + 3)(2r + 5)\]

= \[\displaystyle\sum_{r=1}^{n}4r^2 + 16r + 15\]

= 4\[\displaystyle\sum_{r=1}^{n} r^2 + 16 \displaystyle\sum_{r=1}^{n} r + 15 \displaystyle\sum_{r=1}^{n} 1\]

= `4[(n(n + 1) (2n + 1)) /6] + 16 [(n(n + 1))/2] 15 n`

= `"n"/3[2(2"n"^2 + 3"n" + 1) + 24("n" + 1) + 45]`

= `"n"/3(4"n"^2 + 6"n" + 2 + 24"n" + 24 + 45)`

= `"n"/3(4"n"^2 + 30"n" + 71)`