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Please tell above answer quickly and as soon as possible....................  .....  …
 
7 months ago

Answers : (1)

Samyak Jain
333 Points
							
Repalce n by 1/x for convenience. As n tends to \infty, x tends to 0.
So, given limit is lim { (1+x)m + (1+2x)m + … + (1+kx)m – k } / x
                        x\rightarrow0
Here |x| \rightarrow0, so we can use binomial expansion and neglect higher powers of x.
=  lim { (1+mx+...) + (1+2mx+...) + … + (1+kmx+...) – k } / x
   x\rightarrow0
= lim { (1 + 1 + … k times) + mx(1 + 2 + … + k) … – k } / x
   x\rightarrow0
= lim { k + mx.k(k + 1)/2 – k } / x
   x\rightarrow0
= mk(k + 1) / 2.
7 months ago
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