Advertisements
Advertisements
प्रश्न
Find the sum 22 + 42 + 62 + 82 + ... upto n terms.
Advertisements
उत्तर
22 + 42 + 62 + 82 + ... upto n terms
= (2 x 1)2 + (2 x 2)2 + (2 x 3)2 + (2 + x 4)2 + ...
= \[\displaystyle\sum_{r=1}^{n}(2r^2)\]
= 4\[\displaystyle\sum_{r=1}^{n} r^2\]
= `(4."n"("n" + 1)(2"n" + 1))/6`
= `(2"n"("n" + 1)(2"n" + 1))/3`.
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 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 \[\displaystyle\sum_{r=1}^{n}(5r^2 + 4r - 3)\].
Find \[\displaystyle\sum_{r=1}^{n}\frac{1^2 + 2^2 + 3^2+...+r^2}{2r + 1}\]
Find \[\displaystyle\sum_{r=1}^{n}\frac{1^3 + 2^3 + 3^3 +...+r^3}{(r + 1)^2}\]
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 `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 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 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 \[\displaystyle\sum_{r=1}^{n}\frac{1 + 2 + 3 + ...+ r}{r}\]
Find n, if `(1xx2+2xx3+3xx4+4xx5 + ... + "upto n terms")/(1 + 2 + 3 + 4 + ... + "upto n terms") = 100/3`
Find `\underset{r=1}{\overset{n}{sum}} (1 + 2 + 3 +... + r)/(r)`
