Advertisements
Advertisements
प्रश्न
Find the sum `sum_(r = 1)^n(r + 1)(2r - 1)`.
Advertisements
उत्तर
`sum_(r = 1)^n(r + 1)(2r - 1)`
= `sum_(r = 1)^n(2r^2 + r - 1)`
= `2 sum_(r = 1)^nr^2 + sum_(r = 1)^nr - sum_(r = 1)^n1`
= `2.(n(n + 1)(2n + 1))/6 + (n(n + 1))/2 - "n"`
= `n/6[2(2n^2 + 3n + 1) + 3(n + 1) - 6]`
= `n/6(4n^2 + 6n + 2 + 3n + 3 - 6)`
= `n/6 (4n^2 + 9n - 1)`.
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 (702 – 692) + (682 – 672) + ... + (22 – 12)
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 2 x + 6 + 4 x 9 + 6 x 12 + ... upto n terms.
Find (502 – 492) + (482 –472) + (462 – 452) + .. + (22 –12).
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`
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 \[\displaystyle\sum_{r=1}^{n}\frac{1 + 2 + 3 + ...+ r}{r}\]
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`.
Express the recurring decimal as a rational number.
3.4`bar56`
Find `\underset{r=1}{\overset{n}{sum}} (1 + 2 + 3 +... + r)/(r)`
Find n, if `(1 xx 2 + 2 xx3+3xx4+4xx5+... +"upto n terms")/(1+2+3+4+...+ "upto n terms") = 100/3`.
Find `sum_(r = 1)^n (1 + 2 + 3 + .... + r)/r.`
