Advertisements
Advertisements
प्रश्न
Find 122 + 132 + 142 + 152 + … + 202.
Advertisements
उत्तर
122 + 132 + 142 + 152 + … + 202
= (12 + 22 + 32 + 42 + ... + 202) – (12 + 22 + 32 + 42 + ... + 112)
= \[\displaystyle\sum_{r=1}^{20} r^2 - \displaystyle\sum_{r=1}^{11} r^2\]
= `(20(20 + 1)(2 xx 20 + 1))/6 - (11(11 + 1)(2 xx 11 + 1))/6`
= `(20 xx 21 xx 41)/6 - (11 xx 12 xx 23)/6`
= 2870 – 506 = 2364.
APPEARS IN
संबंधित प्रश्न
Find the sum `sum_(r = 1)^n(r + 1)(2r - 1)`.
Find \[\displaystyle\sum_{r=1}^{n} (3r^2 - 2r + 1)\].
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 \[\displaystyle\sum_{r=1}^{n}r(r-3)(r-2)\].
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 n, if `(1xx2 + 2xx3 + 3xx4 + 4xx5 + .....+ "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 `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`
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+2xx3+3xx4+4xx5+...+"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 . `
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.`
