Question

If \[{{(1+x)}^{n}}={{C}_{0}}+{{C}_{1}}x+{{C}_{2}}{{x}^{2}}+.............+{{C}_{n}}{{x}^{n}}\] then find the value of \[3{{C}_{3}}+8{{C}_{4}}+15{{C}_{5}}+24{{C}_{6}}+......(n-2)\] terms

Answer 1

Hint: We will first differentiate \[{{(1+x)}^{n}}={{C}_{0}}+{{C}_{1}}x+{{C}_{2}}{{x}^{2}}+.............+{{C}_{n}}{{x}^{n}}\] and then divide the whole expression by x and then again differentiate to get the expression which we need to find. Then we will substitute x as 1 in the expression and then we will take the common terms out and then solve to get the answer in terms of n.

Complete step-by-step answer:
It is mentioned in the question that \[{{(1+x)}^{n}}={{C}_{0}}+{{C}_{1}}x+{{C}_{2}}{{x}^{2}}+.............+{{C}_{n}}{{x}^{n}}........(1)\]
So differentiating both sides of equation (1) with respect to x we get,
\[n{{(1+x)}^{n-1}}=0+{{C}_{1}}+2{{C}_{2}}x+3{{C}_{3}}{{x}^{2}}+.............+n{{C}_{n}}{{x}^{n-1}}........(2)\]
Now dividing both sides of equation (2) by x we get,
\[\begin{align}
  & \dfrac{n{{(1+x)}^{n-1}}}{x}=\dfrac{0+{{C}_{1}}+2{{C}_{2}}x+3{{C}_{3}}{{x}^{2}}+.............+n{{C}_{n}}{{x}^{n-1}}}{x} \\
 & \dfrac{n{{(1+x)}^{n-1}}}{x}=\dfrac{{{C}_{1}}}{x}+2{{C}_{2}}+3{{C}_{3}}x+4{{C}_{4}}{{x}^{2}}+5{{C}_{5}}{{x}^{3}}+..........+n{{C}_{n}}{{x}^{n-2}}...........(3) \\
\end{align}\]
Again differentiating equation (3) with respect to x on both sides we get,
\[\dfrac{x\times n(n-1){{(1+x)}^{n-2}}-n{{(1+x)}^{n-1}}}{{{x}^{2}}}=\dfrac{-{{C}_{1}}}{{{x}^{2}}}+0+3{{C}_{3}}+8{{C}_{4}}x+15{{C}_{5}}{{x}^{2}}+..........+n(n-2){{C}_{n}}{{x}^{n-3}}...........(4)\]
Now substituting x as 1 in equation (4) we get,
\[\dfrac{1\times n(n-1){{(1+1)}^{n-2}}-n{{(1+1)}^{n-1}}}{{{1}^{2}}}=\dfrac{-{{C}_{1}}}{{{1}^{2}}}+0+3{{C}_{3}}+8{{C}_{4}}\times 1+15{{C}_{5}}\times {{1}^{2}}+..........+n(n-2){{C}_{n}}{{(1)}^{n-3}}...........(5)\]
Now we know that 1 to the power anything is 1 and hence simplifying equation (5) we get,
\[n(n-1){{2}^{n-2}}-n{{2}^{n-1}}=-{{C}_{1}}+3{{C}_{3}}+8{{C}_{4}}+15{{C}_{5}}+..........+n(n-2){{C}_{n}}...........(7)\]
\[{{C}_{1}}=1\] from equation (1). Now substituting this in equation (7) we get,
\[n(n-1){{2}^{n-2}}-n{{2}^{n-1}}=-1+3{{C}_{3}}+8{{C}_{4}}+15{{C}_{5}}+..........+n(n-2){{C}_{n}}...........(8)\]
Now rearranging and simplifying equation (8) we get,
\[\begin{align}
  & {{2}^{n-2}}[n(n-1)-2n]+1=3{{C}_{3}}+8{{C}_{4}}+15{{C}_{5}}+..........+n(n-2){{C}_{n}} \\
 & 3{{C}_{3}}+8{{C}_{4}}+15{{C}_{5}}+..........+n(n-2){{C}_{n}}={{2}^{n-2}}[{{n}^{2}}-3n]+1 \\
\end{align}\]
Hence the value of \[3{{C}_{3}}+8{{C}_{4}}+15{{C}_{5}}+24{{C}_{6}}+......(n-2)\] terms is \[{{2}^{n-2}}[{{n}^{2}}-3n]+1\].

Note: We should remember the differentiation formulas and also the division rule of differentiation. Division rule of differentiation is \[\dfrac{d}{dx}\left[ \dfrac{f(x)}{g(x)} \right]=\dfrac{g(x)f'(x)-f(x)g'(x)}{{{[g(x)]}^{2}}}\].
We can make a mistake in solving equation (3) where we divide the whole expression by x. Also we need to be careful when we substitute x equal to 1 in equation (4). Also we have to keep in mind that 1 to the power something is 1 always and also differentiation of a constant is zero always.