JEE Main 2022 — Binomial Theorem Question with Solution

Given,

$$C_1+3.2C_2+5.3C_3+$$ ...... upto 10 terms

$$=\frac{\alpha .2^{11}}{2^{\beta }-1}$$ ($$C_0+\frac{C_1}{2}+\frac{C_2}{3}$$ + ..... upto 10 terms)

Now,

L.H.S. :-

$$C_1+3.2C_2+5.3C_3+$$ ...... upto 10 terms

$$=1.1C_1+3.2C_2+5.3C_3+$$ ..... upto 10 terms

$$=\sum _{r=1}^{10}r.(2r-1){}^{10}{C}_r$$

$$=\sum _{r=1}^{10}(2r^2-r).{}^{10}{C}_r$$

$$=2.\sum _{r=1}^{10}{r}^2.{}^{10}{C}_r-\sum _{r=1}^{10}r.{}^n{C}_r$$

[We know, $$\sum _{r=1}^{n}r.{}^n{C}_r=n.2^{n-1}$$

and $$\sum _{r=1}^n{r}^2.{}^n{C}_r=\sum _{r=1}^n(r.{}^n{C}_r).r$$

$$=\sum _{r=1}^n\left(r.\frac{n}{r}.{}^{n-1}{C}_{r-1}\right).r$$

$$=\sum _{r=1}^n\left(n.{}^{n-1}{C}_{r-1}\right).r$$

$$=n\sum _{r=1}^n(r-1+1){}^{n-1}{C}_{r-1}$$

$$=n.\sum _{r=1}^n(r-1).{}^{n-1}{C}_{r-1}+n.\sum _{r=1}^n{}^{n-1}{C}_{r-1}$$

$$=n.(n-1).2^{n-2}+n.2^{n-1}$$]

$$=2\left(n(n-1)2^{n-2}+n.2^{n-1}\right)-n.2^{n-1}$$

Put $$n=10$$

$$=2\left(10.9.2^8+10.2^9\right)-10.2^9$$

$$=45.2^{10}+10.2^{10}-5.2^{10}$$

$$=2^{10}(45+10-5)$$

$$=2^{10}.(50)$$

$$=25.2^{11}$$

R.H.S. :-

$$\frac{\alpha .2^{11}}{2^{\beta }-1}$$ ($$C_0+\frac{C_1}{2}+\frac{C_2}{3}+$$ ..... upto 10 terms)

$$\frac{\alpha .2^{11}}{2^{\beta }-1}$$ ($$\frac{C_0}{1}+\frac{C_1}{2}+\frac{C_2}{3}+$$ ..... upto 10 terms)

$$\frac{\alpha .2^{11}}{2^{\beta }-1}\left(\sum _{r=0}^n\frac{{}^n{C}_r}{r+1}\right)$$

$$=\frac{\alpha .2^{11}}{2^{\beta }-1}\left(\sum _{r=0}^n\frac{{}^{n+1}{C}_{r+1}}{n+1}\right)$$

$$=\frac{\alpha .2^{11}}{2^{\beta }-1}\left(\frac{{}^{n+1}{C}_1+{}^{n+1}{C}_2+....+{}^{n+1}{C}_{n+1}}{n+1}\right)$$

$$=\frac{\alpha .2^{11}}{2^{\beta }-1}\left(\frac{{}^{n+1}{C}_0+{}^{n+1}{C}_2+....+{}^{n+1}{C}_{n+1}-{}^{n+1}{C}_0}{n+1}\right)$$

$$=\frac{\alpha .2^{11}}{2^{\beta }-1}\left(\frac{2^{n+1}-1}{n+1}\right)$$

Putting value of $$n=10$$, we get

$$=\frac{\alpha .2^{11}}{2^{\beta }-1}\left(\frac{2^{11}-1}{11}\right)$$

Using L.H.S. = R.H.S.

$$\Rightarrow 25.2^{11}=\frac{\alpha .2^{11}}{2^{\beta }-1}\left(\frac{2^{11}-1}{11}\right)$$

$$\Rightarrow 25.2^{11}=2^{11}\left(\frac{\alpha }{11}\right)\left(\frac{2^{11}-1}{2^{\beta }-1}\right)$$

By comparing both sides,

$$\frac{\alpha }{11}=25\Rightarrow \alpha =275$$

and $$\frac{2^{11}-1}{2^{\beta }-1}=1$$

$$\Rightarrow 2^{11}=2^{\beta }$$

$$\Rightarrow \beta =11$$

$$\therefore$$ $$\alpha +\beta =275+11=286$$