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 the sum `sum_(r = 1)^n(r + 1)(2r - 1)`.
Find \[\displaystyle\sum_{r=1}^{n} (3r^2 - 2r + 1)\].
Find \[\displaystyle\sum_{r = 1}^{n}\frac{1 + 2 + 3 + ... + r}{r}\]
Find the sum 5 × 7 + 9 × 11 + 13 × 15 + ... upto n terms.
Find (702 – 692) + (682 – 672) + ... + (22 – 12)
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 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 122 + 132 + 142 + 152 + … + 202.
Find (502 – 492) + (482 –472) + (462 – 452) + .. + (22 –12).
Find `sum_(r=1)^n (1+2+3+....+ r)/r`
Find n, if `(1xx2 + 2xx 3 + 3xx4 + 4xx5 + ...+"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 `(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`
