Question

Assertion :The sum of the series $1+(1+2+4)+(4+6+9)+(9+12+16)+....+(361+380+400)$ is $8000$. Reason: ${\sum }_{k=1}^n(k^3-(k-1{)}^3)=n^3$, for any natural number $n$

• 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. Both Assertion and Reason are incorrect

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

$\text{Solve:}$

$\text{Given,}$

$\text{Let sum,}S=1+(1+2+4)+(4+6+9)+\dots \dots (361+380+400)$

$\Rightarrow \text{By observation we find that}$

$1^3-0^3=(1-0)(1^2+0.1+0^2)=1$

$1+2+4=(2^3-1^3)=(2-1)(2^2+1\times 2+1^2)$

similarly,

$({20}^3-{19}^3)=(361+380+400)$

$\Rightarrow S=(1^3-0^3)+(2^3-1^3)+(3^3-2^3)+\dots +({20}^3-{19}^3)$

$\Rightarrow S={20}^3=8000$

Now, ${\sum }_{k=1}^n(k^3-(k-1{)}^3)={\sum }_{k=1}^n(k-(k-1)[k^2+k(k-1)+(k-1{)}^2]$

$={\sum }_{k=1}^n(3k^2-3k+1)$

$=3\sum k^2-3\sum k+\sum 1$

but we have to find for $n$ natural

number

$\Rightarrow {\sum }_{k=1}^n(k^3-(k-1{)}^3)=3\sum n^2-3\sum n+\sum 1$

$=\frac{3n(n+1)(2n+1)}{6}-\frac{3\left(n\right)(n+1)}{2}+n$

$=\frac{n}{2}[2n^2+3n+1-3n-3+2]$

$=n^3$

So, Both assertion and reason are

correct and reason is the correct

explanation for assertion.