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# Please tell above answer quickly and as soon as possible....................  .....  …

Samyak Jain
333 Points
2 years ago
Repalce n by 1/x for convenience. As n tends to $\dpi{80} \infty$, x tends to 0.
So, given limit is lim { (1+x)m + (1+2x)m + … + (1+kx)m – k } / x
x$\dpi{80} \rightarrow$0
Here |x| $\dpi{80} \rightarrow$0, so we can use binomial expansion and neglect higher powers of x.
=  lim { (1+mx+...) + (1+2mx+...) + … + (1+kmx+...) – k } / x
x$\dpi{80} \rightarrow$0
= lim { (1 + 1 + … k times) + mx(1 + 2 + … + k) … – k } / x
x$\dpi{80} \rightarrow$0
= lim { k + mx.k(k + 1)/2 – k } / x
x$\dpi{80} \rightarrow$0
= mk(k + 1) / 2.