Latest IES EC Aptitude Question SOLUTION: 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% o

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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

Asked In IES EC A M ANKIT KALLURAYA (9 years ago)

Unsolved Read Solution (5)

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RAKESH  (9 years ago)

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

Is this solution Helpfull? Yes (1) | No   

RAKESH  (9 years ago)

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

Is this solution Helpfull? Yes | No   

RAKESH  (9 years ago)

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

Is this solution Helpfull? Yes | No   

RAKESH  (9 years ago)

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

Is this solution Helpfull? Yes | No   

RAKESH  (9 years ago)

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

Is this solution Helpfull? Yes | No   

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