Advertisements
Advertisements
प्रश्न
Find \[\displaystyle\sum_{r=1}^{n}(5r^2 + 4r - 3)\].
Advertisements
उत्तर
\[\displaystyle\sum_{r=1}^{n}(5r^2 + 4r - 3)\]
= 5\[\displaystyle\sum_{r=1}^{n}r^2 + 4\displaystyle\sum_{r=1}^{n} r - 3\displaystyle\sum_{r=1}^{n} 1\]
= `5.("n"("n" + 1)(2"n" + 1))/6 + 4.("n"("n" + 1))/2 - 3"n"`
= `"n"/6[5(2"n"^2 + 3"n" + 1) + 12("n" + 1) - 18]`
= `"n"/6(10"n"^2 + 15"n" + 5 + 12"n" + 12 - 18)`
= `"n"/6(10"n"^2 + 27"n" - 1)`.
APPEARS IN
संबंधित प्रश्न
Find the sum `sum_(r = 1)^n(r + 1)(2r - 1)`.
Find `sum_("r" = 1)^"n" (1^3 + 2^3 + ... + "r"^3)/("r"("r" + 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 \[\displaystyle\sum_{r=1}^{n}\frac{1^3 + 2^3 + 3^3 +...+r^3}{(r + 1)^2}\]
Find 2 x + 6 + 4 x 9 + 6 x 12 + ... upto n terms.
Find (502 – 492) + (482 –472) + (462 – 452) + .. + (22 –12).
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 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 `(1xx2+2xx3+3xx4+4xx5+...+"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 `\underset{r=1}{\overset{n}{sum}} (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 xx3+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`
