MCQ
Find the sum of first n terms.
- A$\frac{\text{n}(\text{n}+1)}{2}$
- B$\Big(\frac{\text{n}(\text{n}+1)}{2}\Big)$
- C$\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}$
- D$\Big(\frac{\text{n}(\text{n}+1)}{2}\Big)$
solution:
Sum of first n terms = 1+2+3+4+……+n
$\Rightarrow\Big(\frac{\text{n}}{2}\Big)=\text{(a}+\text{b)}=\Big(\frac{\text{n}}{2}\Big)(\text{1}+\text{n})=\frac{\text{n}(\text{n}+1)}{2}.$
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$\frac{1}{100}$
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The sum of the digits in unit place of all the numbers formed with the help of 3, 4, 5 and 6 taken all at a time is.