Advertisements
Advertisements
Question
Find \[\displaystyle\sum_{r = 1}^{n}\frac{1 + 2 + 3 + ... + r}{r}\]
Advertisements
Solution
\[\displaystyle\sum_{r=1}^{n}\frac{1 + 2 + 3 + ...+ r}{r}\]
= \[\displaystyle\sum_{r=1}^{n}\frac{r(r + 1)}{2r}\]
= \[\frac{1}{2}\displaystyle\sum_{r=1}^{n}(r + 1)\]
`= 1/2 [sum_(r=1)^n r + sum_(r=1)^n 1]`
= `1/2[(n(n + 1))/2 + n]`
= `n/4[(n + 1) + 2]`
= `n/4(n + 3)`
APPEARS IN
RELATED QUESTIONS
Find `sum_("r" = 1)^"n" (1^3 + 2^3 + ... + "r"^3)/("r"("r" + 1)`.
Find n, if `(1 xx 2 + 2 xx 3 + 3 xx 4 + 4 xx 5 + ... + "upto n terms")/(1 + 2 + 3 + 4 + ... + "upto n terms")= 100/3`.
Find \[\displaystyle\sum_{r=1}^{n}r(r-3)(r-2)\].
Find \[\displaystyle\sum_{r=1}^{n}\frac{1^3 + 2^3 + 3^3 +...+r^3}{(r + 1)^2}\]
Find n, if `(1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + ...... + "upto n terms")/(1 + 2 + 3 + 4 + ....+ "upto n terms") = 100/3`
Find n, if `(1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + ...+ "upto n terms")/(1 + 2 + 3 + 4 + ...+ "upto n terms") = 100/3`
Find `sum_(r=1)^n (1 + 2 + 3 + ...+ r)/ r`
Find n, if `(1 xx 2 + 2 xx 3 + 3 xx 4 + 4 xx 5 + ... + "upto n terms")/(1 + 2 + 3 + 4 + ... + "upto n terms") = 100/3`
Find n, if `(1xx2 + 2xx 3 + 3xx4 + 4xx5 + ...+"upto n terms")/(1 + 2 + 3 + 4 + ...+"upto n terms") = 100/3`
Find n, if `(1 xx 2 + 2 xx 3 + 3 xx 4 + 4 xx 5 + ... + "upto n terms")/(1 + 2 + 3 + 4 + ... + "upto n terms") = 100/3`.
Find `sum_(r=1)^n(1+2+3+...+r)/r`
Find `sum_(r=1)^n (1 + 2 + 3 + ... + r)/r`
Find n, if `(1xx2+2xx3+3xx4+4xx5 + ... + "upto n terms")/(1 + 2 + 3 + 4 + ... + "upto n terms") = 100/3`
Find n, if `(1 xx 2 + 2 xx 3 + 3 xx 4 + 4 xx 5 + ... + "upto n terms")/(1 + 2 +3 + 4 + ...+ "upto n terms") = 100/3`.
Express the recurring decimal as a rational number.
3.4`bar56`
Find `sum_(r=1)^n (1+2+3+......+r)/r`
Find `sum _(r=1)^(n) (1 + 2 + 3 + ... + r)/r`
Find `\underset{r=1}{\overset{n}{sum}} (1 + 2 + 3 +... + r)/(r)`
Find n, if `(1 xx 2 + 2 xx 3 + 3 xx 4 + 4 xx 5 + ... + "upto n terms")/ (1 + 2 + 3 + 4 + ... + "upto n terms") = 100/3`
