Advertisements
Advertisements
प्रश्न
Find (502 – 492) + (482 –472) + (462 – 452) + .. + (22 –12).
उत्तर
(502 – 492) + (482 –472) + (462 – 452) + .. + (22 –12).
= (502 + 482 + 462 + ... + 22) – (492 + 472 + 452 + ... + 12)
= \[\displaystyle\sum_{r=1}^{25}(2r)^2 - \displaystyle\sum_{r=1}^{25}(2r - 1)^2\]
= \[\displaystyle\sum_{r=1}^{25} 4r^2 - \displaystyle\sum_{r=1}^{25} (4r^2 - 4r + 1)\]
= \[\displaystyle\sum_{r=1}^{25}[4r^2 - (4r^2 - 4r + 1)]\]
= \[\displaystyle\sum_{r=1}^{25}(4r - 1)\]
= 4\[\displaystyle\sum_{r=1}^{n}r - \displaystyle\sum_{r=1}^{n}1\]
= `4 xx (25(25 + 1))/2 - 25`
= `(4(25)(26))/2 - 25`
= 1300 – 25 = 1275.
APPEARS IN
संबंधित प्रश्न
Find \[\displaystyle\sum_{r=1}^{n}\frac{1 + 2 + 3 + ... + r}{r}\]
Find (702 – 692) + (682 – 672) + ... + (22 – 12)
Find the sum 1 x 3 x 5 + 3 x 5 x 7 + 5 x 7 x 9 + ... + (2n – 1) (2n + 1) (2n + 3)
Find \[\displaystyle\sum_{r=1}^{n}r(r-3)(r-2)\].
Find 122 + 132 + 142 + 152 + … + 202.
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 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + ...... + "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 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`
Express the recurring decimal as a rational number.
3.4`bar56`
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 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`
