Advertisements
Advertisements
प्रश्न
Find \[\displaystyle\sum_{r=1}^{n}\frac{1^3 + 2^3 + 3^3 +...+r^3}{(r + 1)^2}\]
Advertisements
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 `sum_("r" = 1)^"n" (1^3 + 2^3 + ... + "r"^3)/("r"("r" + 1)`.
Find the sum 22 + 42 + 62 + 82 + ... upto n terms.
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`.
If S1, S2 and S3 are the sums of first n natural numbers, their squares and their cubes respectively, then show that: 9S22 = S3(1 + 8S1).
Find \[\displaystyle\sum_{r=1}^{n}\frac{1^2 + 2^2 + 3^2+...+r^2}{2r + 1}\]
Find 122 + 132 + 142 + 152 + … + 202.
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 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 `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`
