Advertisements
Advertisements
प्रश्न
Find \[\displaystyle\sum_{r=1}^{n}\frac{1^3 + 2^3 + 3^3 +...+r^3}{(r + 1)^2}\]
उत्तर
\[\displaystyle\sum_{r=1}^{n}\frac{1^3 + 2^3 + 3^3 +...+r^3}{(r + 1)^2}\]
\[\displaystyle\sum_{r=1}^{n}\frac{r^2(r+1)^2}{4} \times \frac{1}{(r+1)^2}\]
= \[\frac{1}{4}\displaystyle\sum_{r=1}^{n}r^2\]
= `1/4*("n"("n" + 1)(2"n" + 1))/6`
= `("n"("n" + 1)(2"n" + 1))/24`.
APPEARS IN
संबंधित प्रश्न
Find \[\displaystyle\sum_{r=1}^{n} (3r^2 - 2r + 1)\].
Find the sum 5 × 7 + 9 × 11 + 13 × 15 + ... upto n terms.
Find the sum 22 + 42 + 62 + 82 + ... upto n terms.
Find the sum 1 x 3 x 5 + 3 x 5 x 7 + 5 x 7 x 9 + ... + (2n – 1) (2n + 1) (2n + 3)
Find \[\displaystyle\sum_{r=1}^{n}(5r^2 + 4r - 3)\].
Find 122 + 132 + 142 + 152 + … + 202.
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 × 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 `sum_(r=1)^n (1 + 2 + 3 + --- +r)/r`
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`
