plz help

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plz help me how to proceed stepwise in the following questions -
1.what is the sum of the series 1²-2²+3²-4²+...-2008²+2009²
2.sum of the series 1+1/2(1+2)+1/3(1+2+3)+1/4(1+2+3+4)+....upto 20 terms is ??
3.The positive integer n for which 2x2²+3x2³+4x2⁴+...+nx2ⁿ=2ⁿ⁺¹⁰ is??

#Mathematics #Algebra

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4 Answers

1357

Manish Shankar ·2009-04-11 22:03:08

first one

1²-2²+3²-4²......

1²+(3²-2²)+(5²-4²)+(7²-6²)+.......+(2009²-2008²)

1+[1.5+1.9+1.13....(1004) terms]

1+5+9+13.............(1005) terms

sum=(n/2)(2a+(n-1)d)
[(1005)/2][2*1+1004*4]=2019045

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Anuj ·2009-04-11 22:28:43

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mona100 ·2009-04-11 22:42:50

Q 2.
T_n= (n th term)=(1+2+3+...n)/n

T_n=(n(n+1))/n

T_n=(n+1)/2

therefore, S_n=0.5[n(n+1)/2 + n]

solving, we get S_n=(n² +3n)/4

put n=20, we get

sum=115

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Anuj ·2009-04-11 23:17:06

shud nt it be 2 instead of n????????

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