Advertisements
Advertisements
प्रश्न
Find \[\displaystyle\sum_{r=1}^{n}r(r-3)(r-2)\].
उत्तर
\[\displaystyle\sum_{r=1}^{n}r(r-3)(r-2)\]
= \[\displaystyle\sum_{r=1}^{n}(r^3 - 5r^2 + 6r)\].
= \[\displaystyle\sum_{r=1}^{n}r^3 - 5\displaystyle\sum_{r=1}^{n}r^2 + 6\displaystyle\sum_{r=1}^{n}r\]
= `("n"^2("n" + 1)^2)/4 - 5("n"("n" + 1)(2"n" + 1))/6 + 6("n"("n" + 1))/2`
= `("n"("n" + 1))/12[3"n"("n" + 1) - 10(2"n" + 1) + 36]`
= `("n"("n" + 1))/12(3"n"^2 + 3"n" - 20"n" - 10 + 36)`
= `("n"("n" + 1))/12(3"n"^2 - 17"n" + 26)`.
APPEARS IN
संबंधित प्रश्न
Find the sum `sum_("r" = 1)^"n"("r" + 1)(2"r" - 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.
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`.
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 \[\displaystyle\sum_{r=1}^{n}\frac{1^3 + 2^3 + 3^3 +...+r^3}{(r + 1)^2}\]
Find 2 x + 6 + 4 x 9 + 6 x 12 + ... upto n terms.
Find 122 + 132 + 142 + 152 + … + 202.
Find 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 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 \[\displaystyle\sum_{r=1}^{n}\frac{1 + 2 + 3 + ...+ r}{r}\]
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 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 n, if `(1xx2+2xx3+3xx4+4xx5+...+"upto n terms")/(1+2+3+4+...+"upto n terms")=100/3 . `
