Grade 11Algebra
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.
1 Answer
Samyak Jain
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.
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
coefficient of xn+1 in [(1+x)n+m+1 – (1+x)n] is n+m+1Cn+1 .
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