0

Guest

Courses New

Please send the solution of both the questions in the above attached image as soon as possible.

Please send the solution of both the questions in the above attached image as soon as possible. 

Question Image
Grade:11

1 Answers

Samyak Jain
333 Points
5 years ago
​13. Ans: Let S = ∑r=0m n+rCn
S = nCn + n+1Cn + ….. + n+mCn
Coefficient of xn in (1+x)n is nCn , that of xn in (1+x)n+1 is n+1Cn and so on.
\therefore S = coefficient of xn in (1+x)n + (1+x)n+1 + ….. + (1+x)n+m.
Now, (1+x)n + (1+x)n+1 + ….. + (1+x)n+m is a G.P. whose first term is (1+x)n, 
common ratio is (1+x) and number of terms is m+1.
So (1+x)n + (1+x)n+1 + ….. + (1+x)n+m = (1+x)n[(1+x)m+1 – 1] / [(1+x) – 1]
                                                              = [(1+x)n+m+1 – (1+x)n] / x
We have to find coefficient of xn in [(1+x)n+m+1 – (1+x)n] / x i.e.
coefficient of xn+1 in [(1+x)n+m+1 – (1+x)n].
General term of (1+x)n+m+1 is Tr+1 = n+m+1Cr xr.
Putting r=n+1, we get Tn+2 = n+m+1Cn+1 xn+1
\because there is no term of xn+1 in (1+x)n,
coefficient of xn+1 in [(1+x)n+m+1 – (1+x)n] is n+m+1Cn+1 .
\therefore S = ∑r=0m n+rCn = n+m+1Cn+1 .

Think You Can Provide A Better Answer ?

Other Related Questions on Algebra

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More