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Question
Find the sum 5 × 7 + 9 × 11 + 13 × 15 + ... upto n terms
Solution
Each term of this series is a product of two numbers.
The first numbers in the product are 5, 9, 13, .... They form an A.P. whose first term is a = 5 and common difference is d = 4. Hence, the first number in the rth product is
a + (r - 1)d = 5 + (r - 1)4 = 4r + 1
The second number in each product exceeds the first number by 2. Hence the second number in the rth product is 4r + 1 + 2 = 4r + 3.
∴ rth term = t, = (4r + 1)(4r + 3)
= 16r2+ 16r + 3
∴ 5 × 7 + 9 × 11 + 11 × 13 + ... upto n terms
`= sum_("r" = 1)^"n" "t"_"r" = sum_("r" = 1)^"n" (16"r"^2 + 16"r" + 3)`
`= 16 sum_("r" = 1)^"n" "r"^2 + 16 sum_("r" = 1)^"n" "r" + 3 sum_("r" = 1)^"n" 1`
`= 16 * ("n"("n" + 1)("2n" + 1))/6 + 16 * ("n"("n + 1"))/2 + 3"n"`
`= 8/3 *`n(n + 1)(2n + 1) + 8 × n(n + 1) + 3n
= n`[(8("n" + 1)(2"n" + 1))/3 + 8("n" + 1) + 3]`
`= "n"[(8(2"n"^2 + 3"n" + 1) + 24("n" + 1) + 9)/3]`
`= "n"/3 (16"n"^2 + 24"n" + 8 + 24"n" + 24 + 9)`
`= "n"/3(16"n"^2 + 48"n" + 41)`
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