1bhaiya i have given everything correctly.... answers is a
√[1+(1/12)+(1/22)] +√[1+(1/22)+(1/32)]+ .....................+√[1+(1/20072)+(1/20082)] =? a) 2008-(1/2008) b)2009-(1/2008) c)2008-(1/2009) d)2009-(1/2009)
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10 Answers
62kamalendu
could you pleasecheck allthe options
I think you havenot written everything correctly
but the answer is a??
9hint
take any general term as root of 1 + 1/r2 + 1/r+12 and rationalise it to get in form of ancillary series
i think ull get it now
21y this cele ?
1 + 1/r^2 + 1/r+1^2
ist the genral term root of " 1 + 1/r2 + 1/(r+1)2
did u xpand this or wat?
62kamalendu.. i was referring to options b and d.. they are the same.
1then the answer will definately be "d"....will explain the answer after lunch....
1each term.......of the form...√1+1/n2+1/(n+1)2
this is equal to (n2+n+1)/(n2+n)=1+1/n-1/(n+1)
given sum=1Σ2008(1+1/n-1/(n+1))
=2008+((1-1/2) + (1/2-1/3) + ......+(1/2008-1/2009))
=2008+1-1/2009
=2009-1/2009