The correct option is
D Assertion is incorrect but Reason is correct
12+22+32+....+n2=n(n+1)(2n+1)6We will prove it by induction method.
For n=1,
LHS=1
RHS=66=1
Hence, the result is true for n=1.
Let the result is true for k
12+22+32+....+k2=k(k+1)(2k+1)6
We will prove for k+1
12+22+32+....+k2+(k+1)2=(k+1)(k+2)(2k+3)6
Consider, LHS=12+22+32+....+k2+(k+1)2
=(k+1)(k+2)(2k+3)6+(k+1)2
=(k+1)[(k+2)(2k+3)6+(k+1)]
=(k+1)[2k2+11k+126]
=(k+1)(k+4)(2k+3)6
=RHS
Hence, by mathematical induction, the result is true for all natural numbers.
Assertion:
(12−a1)+(22−a2)+...+(n2−an)=13n(n2−1)
12+22+32+....+n2−(a1+a1+a1+....+a1)=13n(n2−1)
Let Sn=a1+a2+a3+...+an
⇒∑n2−Sn=13n(n2−1)
⇒Sn=n(n+1)(2n+1)6−13n(n+1)(n−1)
=n(n+1)6[2n+1−2(n−1)]
Sn=n(n+1)2
which is the sum of first n natural numbers
Sn=1+2+3+.....n=∑n
⇒an=n≠n+1
∴ Assertion (A) is false and Reason (R) is true.