MCQ
If $\sum\text{n}=210,$ then $\sum\text{n}^2=$
- ✓2870
- B2160
- C2970
- Dnone of these.
✓
Answer
Correct option: A.
2870
Given,
$\sum\text{n}=210$
$\Rightarrow\text{n}\Big(\frac{\text{n}+1}{2}\Big)=210$
$\Rightarrow\text{n}^2+\text{n}-420=0$
$\Rightarrow(\text{n}-20)(\text{n}+21)=0$
$\Rightarrow\text{n}=20$ $(\because\ \text{n}>0)$
Now,
$\sum\text{n}^2=\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}$
$\Rightarrow\frac{\text{n}(\text{n}+1)}{2}\times\frac{(2\text{n}+1)}{3}$
$\Rightarrow(210)\times\Big(\frac{41}{3}\Big)$
$\Rightarrow(70)\times(41)$
$\Rightarrow2870$
$\sum\text{n}=210$
$\Rightarrow\text{n}\Big(\frac{\text{n}+1}{2}\Big)=210$
$\Rightarrow\text{n}^2+\text{n}-420=0$
$\Rightarrow(\text{n}-20)(\text{n}+21)=0$
$\Rightarrow\text{n}=20$ $(\because\ \text{n}>0)$
Now,
$\sum\text{n}^2=\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}$
$\Rightarrow\frac{\text{n}(\text{n}+1)}{2}\times\frac{(2\text{n}+1)}{3}$
$\Rightarrow(210)\times\Big(\frac{41}{3}\Big)$
$\Rightarrow(70)\times(41)$
$\Rightarrow2870$
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