Sequences and Series

We have to check each option, From option (a), we have,
(2n+7)<(n+3)2
Let P(n)≡(2n+7)<(n+3)2
P(1)=(2+7)=9<(1+3)2=42=16
Let us assume that P(K) is true.
We have to prove that P(K+1) is true
‌(2K+7)<(K+3)2
‌2K+7<K2+9+6K
Add 2 on LHS and 7+2K on RHS
‌ as ‌2<7+2K
⇒2K+9<K2+16+8K
2(K+1)+7<((K+1)+3)2
∴P(K+1) is true.
Hence, P(x) is true, ∀n∈N.
From option b,
Let Q(n)≡12+22+...+n2>‌

n3

3

Q(1)=12>‌

1

3

Let Q(K) be true.
12+22+32+...+K2>‌

K3

3

12+22+32+...+K2
+(K+1)2>‌

K3

3

+(K+1)2
12+22+32+...+K2+(K+1)2
>‌

K3+3K2+6K+3

3

12+22+32+...+K2+(K+1)2>
‌

K3+3K2+3K+1+3K+2

3

=‌

(K+1)3+3K+2

3

>‌

(K+1)3

3

∴Q(K+1)2 is true.
Hence, Q(n) is true, ∀n∈N
From option (c),
R(n)=352n+1+23n+1

R(1)=3×125+24=375+16=391
which is divisible by 23
Let us assume it is true for n=k,k∈N
∴S(k)=3⋅52k+1+23k+1 is divisible by
23‌∀k∈N
Let 3⋅52k+1+23k+1=23t,t∈N
∴3⋅5k2k+1=23t−23k+1
Now, we have to prove that for n=k+1
∴S(k+1)‌=3⋅52(k+1)+1+23(k+1)+1
‌=3⋅52k+1⋅52+23k+1⋅23
‌=(23t−23k+1)25+8⋅23k+1
‌=23t×25−25×23k+1+8⋅23k+1
∴S(k+1)=23t×25−1723k+1
Hence, 3⋅52n+1+23n+1 is not divisible by 23 .
Hence, R(n) is not divisible by 23,∀n∈N. From option (d),
2+7+12+...+(5n−3) is an AP with common difference as 5.
Sn=‌

n

2

[2×2+(n−1)5]
=‌

n

2

(5n−1)