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Question
Prove by method of induction, for all n ∈ N:
`1/(3.5) + 1/(5.7) + 1/(7.9) + ...` to n terms = `"n"/(3(2"n" + 3))`
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Solution
Let P(n) ≡ `1/(3.5) + 1/(5.7) + 1/(7.9) + ...` to n terms = `"n"/(3(2"n" + 3))`, for all n ∈ N
But the first factor in each term of the denominator
i.e., 3, 5, 7, … are in A.P. with a = 3 and d = 2.
∴ nth term = a + (n – 1)d = 3 + (n – 1)2 = (2n + 1)
Also, second factor in each term of denominator
i.e., 5, 7, 9, … are in A.P. with a = 5 and d = 2.
∴ nth term = a + (n – 1)d = 5 + (n – 1)2 = (2n + 3)
∴ nth term tn = `1/((2"n" + 1)(2"n" + 3))`
∴ P(n) ≡ `1/(3.5) + 1/(5.7) + 1/(7.9) + ... + 1/((2"n" + 1)(2"n" + 3)) = "n"/(3(2"n" + 3))`
Step I:
Put n = 1
L.H.S. = `1/(3.5) = 1/15`
R.H.S. = `1/(3[2(1) + 3]) = 1/(3(2 + 3)) = 1/15` = L.H.S.
∴ P(n) is true for n = 1
Step II:
Let us consider that P(n) is true for n = k
∴ `1/(3.5) + 1/(5.7) + 1/(7.9) + ... + 1/((2"k" + 1).(2"k" + 3)) = "k"/(3(2"k" + 3))` ...(i)
Step III:
We have to prove that P(n) is true for n = k + 1
i.e., to prove that
`1/(3.5) + 1/(5.7) + 1/(7.9) + ... + 1/((2"k" + 3).(2"k" + 5)) = ("k" + 1)/(3(2"k" + 5))`
L.H.S. = `1/(3.5) + 1/(5.7) + 1/(7.9) + ... + 1/((2"k" + 3)(2"k" + 5))`
= `1/(3.5) + 1/(5.7) + 1/(7.9) + ... + 1/((2"k" + 1)(2"k" + 3)) + 1/((2"k" + 3)(2"k" + 5))`
= `"k"/(3(2"k" + 3)) + 1/((2"k" + 3)(2"k" + 5))` ...[From (i)]
= `("k"(2"k" + 5) + 3)/(3(2"k" + 3)(2"k" + 5))`
= `(2"k"^2 + 5"k" + 3)/(3(2"k" + 3)(2"k" + 5))`
= `(2"k"^2 + 2"k" + 3"k" + 3)/(3(2"k" + 3)(2"k" + 5)`
= `(2"k"("k" + 1) + 3("k" + 1))/(3(2"k" + 3)(2"k" + 5))`
= `((2"k" + 3)("k" + 1))/(3("2k" + 3)(2"k" + 5))`
= `("k" + 1)/(3(2"k" + 5))`
= R.H.S.
∴ P(n) is true for n = k + 1
Step IV:
From all steps above by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ `1/(3.5) + 1/(5.7) + 1/(7.9) + ...` to n terms = `"n"/(3(2"n" + 3))` for all n ∈ N.
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