plz answer these questions?

Question

Q.1 1+2+3+4+..........n= n/2(n+1),
2+4+6+*+..........2n= n(n+1),
1+3+5+ ...............(2n+1) or (2n-1)?????

nth term = 1st term + (n-1)*comman difference
Tn =1+(n-1)*2 =(2n-1)
so, series is 1+3+5+ ...............(2n-1)
formula of sum of above series =(n/2)*(1+(2n-1)) =n^2

Q.2 1+4+9+16+...........n square= 1/6n(n+1)(2n+1),
2sq+3sq+4sq..........2n sq= 2/3n(n+1)(2n+1),

1+3sq+5 sq+............(2n-1)sq or (2n+1)sq ????????????

1+3sq+5 sq+............(2n-1)sq

sum explanation

1sq+2sq+3sq+4sq+............(2n-1)sq+(2n)sq=[2n(2n+1)(4n+1)]/6
combine odd and even terms
[1sq+3sq+5sq+........(2n-1)sq]+[2sq+4sq+......(2n)sq]=[2n(2n+1)(4n+1)]/6
=>[1sq+3sq+5sq+........(2n-1)sq]=[2n(2n+1)(4n+1)]/6 - 2/3n(n+1)(2n+1)

Q.3 1cube+2 cube++3 cube+............n cube,
2 cube+4 cube+ 6 cube+............2n cube= 2n sq(n+1)sq,
1+3 cube+5 cube+...............(2n-1)cube or (2n+1)cube??????????
on same way as sq[/b]

ajith · Answer

nitinrajasingh wrote:Q.1 1+2+3+4+..........n= n/2(n+1),
2+4+6+*+..........2n= n(n+1),
1+3+5+ ...............(2n+1) or (2n-1)?????
Q.2 1+4+9+16+...........n square= 1/6n(n+1)(2n+1),
2sq+3sq+4sq..........2n sq= 2/3n(n+1)(2n+1),
1+3cube+5 cube+............(2n-1)sq or (2n+1)sq ????????????
Q.3 1cube+2 cube++3 cube+............n cube,
2 cube+4 cube+ 6 cube+............2n cube= 2n sq(n+1)sq,
1+3 cube+5 cube+...............(2n-1)cube or (2n+1)cube??????????
Q.4 Sum of all prime no. upto n=????(formula ?),
Sum of square of all prime no. upto n= formula?????
Sum of cube of all prime no. upto n= formula?????
Q.5 Sum of all composite no. upto n=????(formula ?),
Sum of square of all composite no. upto n= formula?????
Sum of cube of all composite no. upto n= formula?????

I am sure 4 and 5 have no formulas

1. 1+3+5+ ...............(2n-1) = n^2
2.&3 You can derive the formula based on Total sum = Sum of Evens + Sum of Odds
so Sum of odds = Total Sum - Sum of evens ( careful about n here)

sanjayism · Answer

Q.1 1+2+3+4+..........n= n/2(n+1),
2+4+6+*+..........2n= n(n+1),
1+3+5+ ...............(2n+1) or (2n-1)?????

nth term = 1st term + (n-1)*comman difference
Tn =1+(n-1)*2 =(2n-1)
so, series is 1+3+5+ ...............(2n-1)
formula of sum of above series =(n/2)*(1+(2n-1)) =n^2

Q.2 1+4+9+16+...........n square= 1/6n(n+1)(2n+1),
2sq+3sq+4sq..........2n sq= 2/3n(n+1)(2n+1),

1+3sq+5 sq+............(2n-1)sq or (2n+1)sq ????????????

1+3sq+5 sq+............(2n-1)sq

sum explanation

1sq+2sq+3sq+4sq+............(2n-1)sq+(2n)sq=[2n(2n+1)(4n+1)]/6
combine odd and even terms
[1sq+3sq+5sq+........(2n-1)sq]+[2sq+4sq+......(2n)sq]=[2n(2n+1)(4n+1)]/6
=>[1sq+3sq+5sq+........(2n-1)sq]=[2n(2n+1)(4n+1)]/6 - 2/3n(n+1)(2n+1)

Q.3 1cube+2 cube++3 cube+............n cube,
2 cube+4 cube+ 6 cube+............2n cube= 2n sq(n+1)sq,
1+3 cube+5 cube+...............(2n-1)cube or (2n+1)cube??????????
on same way as sq[/b]

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