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=‌