Answer: (B) = \(\frac{n(4n^2+6n-1)}{3}\)
Let Sn = 1.3 + 3.5 + 5.7 + .... + (2n – 1) (2n + 1)
When n = 1, Sn = 1.3 = 3
From the given options, when n = 1,
\(\frac{n(2n^2+3n+1)}{6} =\frac{1\times(2+3+1)}{6}\) =1 ≠ Sn
\(\frac{n(4n^2+6n-1)}{3} = \frac{1\times (4+6-1)}{3} =\frac{9}{3}\) = 3 = Sn
\(\frac{n(n^2+4)}{6} = \frac{1\times 5}{6} =\frac{5}{6}
\) ≠ Sn
\(\frac{n^2(4n^2+5)}{3} = \frac{1\times (4+5)}{3} =\frac{9}{3} =3 =S_n\)
When n = 2, Sn = 1.3 + 3.5 = 3 + 15 = 18
\(\frac{n(4n^2+6n-1)}{3} = \frac{2 (4\times 4+12-1)}{3} =18=S_n\)
\(\frac{n^2(4n^2+5)}{3} = \frac{4\times (4\times 4+5)}{3} =\frac{4\times 21}{3}\) = 28 ≠ Sn. .
∴ Sn =\(\frac{n}{3}\) (4n2+6n–1) for n = 1 and n = 2
Now we shall show that Sn = \(\frac{n}{3}\)(4n2 + 6n – 1) for all n∈N.
Using the principle of mathematical induction.
Let T(n) = 1.3 + 3.5 + 5.7 + .... + (2n – 1) (2n + 1)
= \(\frac{n}{3}\) (4n2 + 6n – 1)
Basic Step:
For n = 1, LHS = 1.3 = 3
RHS = \(\frac{1\times (4+6-1)}{3} = \frac{9}{3}\) = 3.
∴ LHS = RHS ⇒ T(1) is true.
Induction Step:
Assume T(n) to be true for n = k, k∈N
1.3 + 3.5 + 5.7 + .... + (2k–1) (2k+1) = \(\frac{k}{3}\) (4k2 + 6k – 1)
To obtain T(k + 1),
add the (k + 1)th term = [(2(k+1)–1)(2(k+1)+1)]
= (2k+1) (2k+3) to both the sides of T(k), i.e.,
1.3 + 3.5 + 5.7 + .... + (2k–1)(2k+1)+(2(k+1)–1)(2(k+1)+1)
=\(\frac{1}{3}\)k(4k2+6k–1) + (2k+1)(2k+3)
= \(\frac{1}{3}\)k(4k2+6k–1) + (4k2+8k+3)
= \(\frac{1}{3}\) {4k3+6k2–k+12k2+24k+9}
= \(\frac{1}{3}\) {4k3+18k2+23k+9}
= \(\frac{1}{3}\) (k+1)(4k2+14k+9)
= \(\frac{1}{3}\) (k+1) {4(k+1)2 + 6(k+1)–1}
∴ T (k + 1) is true, whenever T(k) is true.
Hence, by the principle of mathematical induction, T(n) is true for all n∈N.
⇒ 1.3 + 3.5 + 5. 7 + ..... + (2n–1) (2n+1)
= \(\frac{1}{3}\) n (4n2+6n–1) is true ∀ n∈N.
