We have to check each option, From option (a), we have,
(2n+7)<(n+3)2Let
P(n)≡(2n+7)<(n+3)2P(1)=(2+7)=9<(1+3)2=42=16Let us assume that
P(K) is true.
We have to prove that
P(K+1) is true
‌(2K+7)<(K+3)2‌2K+7<K2+9+6KAdd 2 on LHS and
7+2K on RHS
‌ as ‌2<7+2K⇒2K+9<K2+16+8K2(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>‌Q(1)=12>‌Let
Q(K) be true.
12+22+32+...+K2>‌12+22+32+...+K2+(K+1)2>‌+(K+1)212+22+32+...+K2+(K+1)2>‌12+22+32+...+K2+(K+1)2>‌=‌>‌∴Q(K+1)2 is true.
Hence,
Q(n) is true,
∀n∈NFrom option (c),
R(n)=352n+1+23n+1R(1)=3×125+24=375+16=391which 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∈NLet
3⋅52k+1+23k+1=23t,t∈N∴3⋅5k2k+1=23t−23k+1Now, 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+1Hence,
3â‹…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=‌[2×2+(n−1)5]=‌(5n−1)