Please tell above answer quickly and as soon as p

Question

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

Samyak Jain · Accepted Answer

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 $\rightarrow$ 0

Here |x| $\rightarrow$ 0, so we can use binomial expansion and neglect higher powers of x.

= lim { (1+mx+...) + (1+2mx+...) + … + (1+kmx+...) – k } / x

x $\rightarrow$ 0

= lim { (1 + 1 + … k times) + mx(1 + 2 + … + k) … – k } / x

x $\rightarrow$ 0

= lim { k + mx.k(k + 1)/2 – k } / x
x $\rightarrow$ 0

= mk(k + 1) / 2.

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