Advertisements
Advertisements
प्रश्न
Find the sum `sum_("r" = 1)^"n"("r" + 1)(2"r" - 1)`.
Advertisements
उत्तर
`sum_("r" = 1)^"n"("r" + 1)(2"r" - 1)`
= `sum_("r" = 1)^"n"(2"r"^2 + "r" - 1)`
= `2 sum_("r" = 1)^"n""r"^2 + sum_("r" = 1)^"n""r" - sum_("r" = 1)^"n"1`
= `2.("n"("n" + 1)(2"n" + 1))/6 + ("n"("n" + 1))/2 - "n"`
= `"n"/6[2(2"n"^2 + 3"n" + 1) + 3("n" + 1) - 6]`
= `"n"/6(4"n"^2 + 6"n" + 2 + 3"n" + 3 - 6)`
= `"n"/6 (4"n"^2 + 9"n" - 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.
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 \[\displaystyle\sum_{r=1}^{n}(5r^2 + 4r - 3)\].
Find \[\displaystyle\sum_{r=1}^{n}r(r-3)(r-2)\].
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 n, if `(1xx2 + 2xx3 + 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 `(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 `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`
