Advertisements
Advertisements
Question
Find the sum 22 + 42 + 62 + 82 + ... upto n terms.
Advertisements
Solution
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
RELATED QUESTIONS
Find \[\displaystyle\sum_{r = 1}^{n}\frac{1 + 2 + 3 + ... + r}{r}\]
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 \[\displaystyle\sum_{r=1}^{n}\frac{1^2 + 2^2 + 3^2+...+r^2}{2r + 1}\]
Find \[\displaystyle\sum_{r=1}^{n}\frac{1^3 + 2^3 + 3^3 +...+r^3}{(r + 1)^2}\]
Find 2 x + 6 + 4 x 9 + 6 x 12 + ... upto n terms.
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 `(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 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 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.`
