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Numerical Ability
Algebra
Below given are two formulae for long summation.
1) 1 + 2 + 3 + 4 + ................ + n = n(n+1)/2
2) 1^2 + 2^2 + 3^2 + 4^2 + ........ + n^2 = 1/6 n(n+1)(2n+1)
The value of
(1% of 2 + 2% of 3 + 3% of 4 + ........ + 99% of 100) will be
Read Solution (Total 5)
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- 1% of 2 + 2% of 3 + 3% of 4 + ........ + 99% of 100
= ( 1* 2 + 2* 3 + 3* 4 + ........ + 99 *100) / 100
= (1/100)* [1(1+1) + 2(1+2) + 3(1+3) + ......... + 99(1+99)]
= (1/100) * [(1+2+3+....+99) + (1^2 + 2^2 +3^2 +.....+ 99^2]
= 1/100 * [ n(n+1)/2 + 1/6 n(n+1)(2n+1) ] put n=99
= 3333 - 10 years agoHelpfull: Yes(1) No(0)
- 1% of 2 + 2% of 3 + 3% of 4 + ........ + 99% of 100
= ( 1* 2 + 2* 3 + 3* 4 + ........ + 99 *100) / 100
= (1/100)* [1(1+1) + 2(1+2) + 3(1+3) + ......... + 99(1+99)]
= (1/100) * [(1+2+3+....+99) + (1^2 + 2^2 +3^2 +.....+ 99^2]
= 1/100 * [ n(n+1)/2 + 1/6 n(n+1)(2n+1) ] put n=99
= 3333 - 10 years agoHelpfull: Yes(0) No(0)
- 1% of 2 + 2% of 3 + 3% of 4 + ........ + 99% of 100
= ( 1* 2 + 2* 3 + 3* 4 + ........ + 99 *100) / 100
= (1/100)* [1(1+1) + 2(1+2) + 3(1+3) + ......... + 99(1+99)]
= (1/100) * [(1+2+3+....+99) + (1^2 + 2^2 +3^2 +.....+ 99^2]
= 1/100 * [ n(n+1)/2 + 1/6 n(n+1)(2n+1) ] put n=99
= 3333 - 10 years agoHelpfull: Yes(0) No(0)
- 1% of 2 + 2% of 3 + 3% of 4 + ........ + 99% of 100
= ( 1* 2 + 2* 3 + 3* 4 + ........ + 99 *100) / 100
= (1/100)* [1(1+1) + 2(1+2) + 3(1+3) + ......... + 99(1+99)]
= (1/100) * [(1+2+3+....+99) + (1^2 + 2^2 +3^2 +.....+ 99^2]
= 1/100 * [ n(n+1)/2 + 1/6 n(n+1)(2n+1) ] put n=99
= 3333 - 10 years agoHelpfull: Yes(0) No(0)
- 1% of 2 + 2% of 3 + 3% of 4 + ........ + 99% of 100
= ( 1* 2 + 2* 3 + 3* 4 + ........ + 99 *100) / 100
= (1/100)* [1(1+1) + 2(1+2) + 3(1+3) + ......... + 99(1+99)]
= (1/100) * [(1+2+3+....+99) + (1^2 + 2^2 +3^2 +.....+ 99^2]
= 1/100 * [ n(n+1)/2 + 1/6 n(n+1)(2n+1) ] put n=99
= 3333 - 10 years agoHelpfull: Yes(0) No(0)
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