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Question: Let \({S_1} = \sum\limits_{j = 1}^{10} {j\left( {j - 1} \right)} {}^{10}{C_j},{S_2} = \sum\limits_{j...

Let S1=∑j=110j(j−1)10Cj,S2=∑j=110j10Cj{S_1} = \sum\limits_{j = 1}^{10} {j\left( {j - 1} \right)} {}^{10}{C_j},{S_2} = \sum\limits_{j = 1}^{10} {j{}^{10}{C_j}} andS3=∑j=110j210Cj{S_3} = \sum\limits_{j = 1}^{10} {{j^2}{}^{10}{C_j}} .
Statement-1: S3=55×29{S_3} = 55 \times {2^9}
Statement-2: S1=90×28{S_1} = 90 \times {2^8} and S2=10×28{S_2} = 10 \times {2^8}
(A) Statement - 1 is true, Statement - 2 is true; Statement - 2 is not the correct explanation for Statement - 1
(B) Statement - 1 is true, Statement - 2 is false
(C) Statement - 1 is false, Statement - 2 is true
(D) Statement - 1 is true, Statement - 2 is true; Statement - 2 is the correct explanation for Statement – 1

Explanation

Solution

Try to solve each of the summations one by one. Start with S1{S_1}, and use the property nCr=nrn−1Cr−1=n(n−1)r(r−1)n−2Cr−2{}^n{C_r} = \dfrac{n}{r}{}^{n - 1}{C_{r - 1}} = \dfrac{{n\left( {n - 1} \right)}}{{r\left( {r - 1} \right)}}{}^{n - 2}{C_{r - 2}} on it to simplify it. Then use nC0+nC1+nC2+..........+nCn=2n{}^n{C_0} + {}^n{C_1} + {}^n{C_2} + .......... + {}^n{C_n} = {2^n} to get the final value of S1{S_1}. Similarly, solve S2{S_2} using the above two properties. In S3{S_3}, use j2=(j(j−1)+j){j^2} = \left( {j\left( {j - 1} \right) + j} \right) to split the summation into two parts.

Complete step-by-step answer:
Here we have three summations S1,S2{S_1},{S_2}andS3{S_3} to be solved. Let’s go step by step while solving each of them one by one.
We can use the property of combination that says:nCr=nrn−1Cr−1=n(n−1)r(r−1)n−2Cr−2{}^n{C_r} = \dfrac{n}{r}{}^{n - 1}{C_{r - 1}} = \dfrac{{n\left( {n - 1} \right)}}{{r\left( {r - 1} \right)}}{}^{n - 2}{C_{r - 2}}, in S1{S_1}
⇒S1=∑j=110j(j−1)10Cj=∑j=110j(j−1)×10(10−1)(j−1)×8Cj−2\Rightarrow {S_1} = \sum\limits_{j = 1}^{10} {j\left( {j - 1} \right)} {}^{10}{C_j} = \sum\limits_{j = 1}^{10} {j\left( {j - 1} \right)} \times \dfrac{{10\left( {10 - 1} \right)}}{{\left( {j - 1} \right)}} \times {}^8{C_{j - 2}}
So, now it can be rewritten as:
⇒S1=∑j=110j(j−1)×10(10−1)j(j−1)×8Cj−2=9×10×∑j=1108Cj−2\Rightarrow {S_1} = \sum\limits_{j = 1}^{10} {j\left( {j - 1} \right)} \times \dfrac{{10\left( {10 - 1} \right)}}{{j\left( {j - 1} \right)}} \times {}^8{C_{j - 2}} = 9 \times 10 \times \sum\limits_{j = 1}^{10} {{}^8{C_{j - 2}}}
Also, we know the property: nC0+nC1+nC2+..........+nCn=2n{}^n{C_0} + {}^n{C_1} + {}^n{C_2} + .......... + {}^n{C_n} = {2^n}
⇒S1=9×10×∑j=1108Cj−2=9×10×28=90×28\Rightarrow {S_1} = 9 \times 10 \times \sum\limits_{j = 1}^{10} {{}^8{C_{j - 2}}} = 9 \times 10 \times {2^8} = 90 \times {2^8}
Therefore, we got S1=90×28{S_1} = 90 \times {2^8} (1)
Now, let’s move to S2{S_2}
Again using the propertynCr=nrn−1Cr−1=n(n−1)r(r−1)n−2Cr−2{}^n{C_r} = \dfrac{n}{r}{}^{n - 1}{C_{r - 1}} = \dfrac{{n\left( {n - 1} \right)}}{{r\left( {r - 1} \right)}}{}^{n - 2}{C_{r - 2}}, in S2{S_2}
⇒S2=∑j=110j10Cj=∑j=110j×10j×9Cj−1=10×∑j=1109Cj−1\Rightarrow {S_2} = \sum\limits_{j = 1}^{10} {j{}^{10}{C_j}} = \sum\limits_{j = 1}^{10} {j \times \dfrac{{10}}{j} \times {}^9{C_{j - 1}}} = 10 \times \sum\limits_{j = 1}^{10} {{}^9{C_{j - 1}}}
Here, we can now use: nC0+nC1+nC2+..........+nCn=2n{}^n{C_0} + {}^n{C_1} + {}^n{C_2} + .......... + {}^n{C_n} = {2^n}

Therefore, we get $${S_2} = 10 \times {2^9}$$ (2) Now, we move on to ${S_3}$ We can change ${j^2} = \left( {j\left( {j - 1} \right) + j} \right)$ in${S_3}$, we get: $$ \Rightarrow {S_3} = \sum\limits_{j = 1}^{10} {{j^2}{}^{10}{C_j}} = \sum\limits_{j = 1}^{10} {\left( {j\left( {j - 1} \right) + j} \right) \times {}^{10}{C_j}} $$ Now, we can easily separate the above expression in two summations as: $$ \Rightarrow {S_3} = \sum\limits_{j = 1}^{10} {\left( {j\left( {j - 1} \right) + j} \right) \times {}^{10}{C_j}} = \sum\limits_{j = 1}^{10} {j\left( {j - 1} \right) \times {}^{10}{C_j}} + \sum\limits_{j = 1}^{10} {j \times {}^{10}{C_j}} $$ This can be written as, by using${S_1} = \sum\limits_{j = 1}^{10} {j\left( {j - 1} \right)} {}^{10}{C_j},{S_2} = \sum\limits_{j = 1}^{10} {j{}^{10}{C_j}} $: $$ \Rightarrow {S_3} = \sum\limits_{j = 1}^{10} {j\left( {j - 1} \right) \times {}^{10}{C_j}} + \sum\limits_{j = 1}^{10} {j \times {}^{10}{C_j}} = {S_1} + {S_2}$$ Now, we use the relation (1) and (2) in the above equation: $$ \Rightarrow {S_3} = {S_1} + {S_2} = \left( {90 \times {2^8}} \right) + \left( {10 \times {2^9}} \right) = \left( {45 + 10} \right) \times {2^9} = 55 \times {2^9}$$ Hence, we found ${S_1} = 90 \times {2^8}$,$${S_2} = 10 \times {2^9}$$and $${S_3} = 55 \times {2^9}$$ So, we can say Statement-1 is correct but Statement-2 is false **Thus, the option (B) is the correct option.** **Note:** Try to go step by step in the solution with each of the summation one by one. Here combination ${}^n{C_r}$ can be defined as $\dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ where $n!$ represents the factorial of $n$ , i.e. the product of natural number till $n$. An alternative approach for this problem can be to start with the summation ${S_3}$ first and transform it to form the relation $${S_3} = {S_1} + {S_2}$$. And then solve $${S_1}$$ and $${S_2}$$