Question

Assertion :If $(1^2-a_1)+(2^2-a_2)+...+(n^2-a_n)=\frac{1}{3}n(n^2-1)$ then $a_n=n+1$ Reason: $1^2+2^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is D Assertion is incorrect but Reason is correct

$1^2+2^2+3^2+....+n^2=\frac{n(n+1)(2n+1)}{6}$

We will prove it by induction method.

For $n=1$,

$LHS=1$

$RHS=\frac{6}{6}=1$

Hence, the result is true for $n=1$.
Let the result is true for $k$

$1^2+2^2+3^2+....+k^2=\frac{k(k+1)(2k+1)}{6}$

We will prove for $k+1$

$1^2+2^2+3^2+....+k^2+(k+1{)}^2=\frac{(k+1)(k+2)(2k+3)}{6}$
Consider, $LHS=1^2+2^2+3^2+....+k^2+(k+1{)}^2$

$=\frac{(k+1)(k+2)(2k+3)}{6}+(k+1{)}^2$

$=(k+1)[\frac{(k+2)(2k+3)}{6}+(k+1\left)\right]$

$=(k+1)\left[\frac{2k^2+11k+12}{6}\right]$

$=(k+1)\frac{(k+4)(2k+3)}{6}$

$=RHS$

Hence, by mathematical induction, the result is true for all natural numbers.
Assertion:

$(1^2-a_1)+(2^2-a_2)+...+(n^2-a_n)=\frac{1}{3}n(n^2-1)$
$1^2+2^2+3^2+....+n^2-(a_1+a_1+a_1+....+a_1)=\frac{1}{3}n(n^2-1)$
Let $S_n=a_1+a_2+a_3+...+a_n$

$\Rightarrow \sum n^2-S_n=\frac{1}{3}n(n^2-1)$
$\Rightarrow S_n=\frac{n(n+1)(2n+1)}{6}-\frac{1}{3}n(n+1)(n-1)$
$=\frac{n(n+1)}{6}[2n+1-2(n-1)]$
$S_n=\frac{n(n+1)}{2}$

which is the sum of first n natural numbers

$S_n=1+2+3+.....n=\sum n$

$\Rightarrow a_n=n\ne n+1$

$\therefore$ Assertion (A) is false and Reason (R) is true.