JEE Main 2022 · 30 Jun · Shift 1 · Mathematics Q7

MathematicsSingle correct+4 / -1Question 7 of 90

The value of $1+\dfrac1{1+2}+\dfrac1{1+2+3}+\dots+\dfrac1{1+2+3+\dots+11}$ is equal to :

\dfrac{20}{11}

\dfrac{11}6

\dfrac{241}{132}

\dfrac{21}{11}

Answer and worked solution

Official answer: B

Terms. $\dfrac2{n(n+1)}=2\left(\frac1n-\frac1{n+1}\right)$ ; summed to 11: $2\left(1-\frac1{12}\right)=\frac{11}6$ .

(B)

Topics:Special series sums

JEE Main 2022 · 30 Jun · Shift 1, Mathematics question 7