# Polynomials

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## Sequences

2 + 6 + 10 + 14 ………..upto n term is given by Sn. How many of the following statements are true?
1. S2m – S2k could be a multiple of 16
2. 18Sn is a perfect square for all n
3. S2n > 2Sn for all n > 1
4. Sm+n > Sm + Sn for all m, n > 1

1. 1
2. 2
3. 3
4. 4

Choice D. 4

## Detailed Solution

Sn = 2 (1 + 3 + 5 + …+ n terms)
tn = 2 + (n – 1)4 = 4n – 2
${S_n = \frac{n}{2} * (2 + 4n – 2) = \frac{(4 * n * n)}{2} = 2*n*n = 2n^2}$

1. S2m – S2k = (2 * 2m * 2m) – (2 * 2k * 2k) = 8 * (m – k) (m + k)
This could be a multiple of 16 if both m and k are odd or if both m and k are even.

2. 18Sn = 18 × 2 × n × n = 36 × n × n
Hence, it is always a perfect square.

3. S2n = 2 × 2n × 2n = 8 × n × n = 8n2
2Sn = 2 × 2 × n × n = 4 × n × n = 4n2
Hence, S2n > 2Sn for all n > 1

4. Sm+n = 2 × (m + n) × (m + n)
= 2 × (m2 + n2 + 2mn) = 2m2 + 2n2 + 4mn
On the other hand,
Sm + Sn = 2m2 + 2n2

Hence, Sm+n > Sm + Sn for all m, n > 1

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