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1

WB JEE 2009

Subjective

Show that, for a positive integer n, the coefficient of xk (0 $$\le$$ k $$\le$$ n) in the expansion of 1 + (1 + x) + (1 + x)2 + ....... + (1 + x)n is $${}^{n + 1}{C_{n - k}}$$.

Answer

.

Explanation

$$1 = (1 + x) + {(1 + x)^2} + \,\,......\,\, + {(1 + x)^n}$$

$$ = {{{{(1 + x)}^{n + 1}} - 1} \over {(1 + x) - 1}}$$ ($$\because$$ $${S_n} = {{a({r^n} - 1)} \over {r - 1}}$$)

$$ = {1 \over x}[1 + {}^{n + 1}{C_0}x + {}^{n + 1}{C_1}{x^2} + \,\,.....\,\, + {}^{n + 1}{C_{k + 1}}{x^{k + 1}} + \,\,.....\,\, + {x^{n + 1}}]$$

$$={1 \over x}[{}^{n + 1}{C_0}x + {}^{n + 1}{C_1}{x^2} + \,\,......\,\, + {}^{n + 1}{C_{k + 1}}{x^{k + 1}} + \,\,.....\,\, + \,{x^{n + 1}}]$$

$$ = {}^{n + 1}{C_0} + {}^{n + 1}{C_1}x + \,\,.....\,\, + {}^{n + 1}{C_{k + 1}}{x^k} + \,\,.....\,\, + {x^n}$$

$$\therefore$$ Coefficient of xk is $${}^{n + 1}{C_{k + 1}} = {}^{n + 1}{C_{n - k}}$$ ($$\because$$ $${}^n{C_r} = {}^n{C_{n - r}}$$)

2

WB JEE 2008

Subjective

$$A = \left[ {\matrix{ 1 & 2 \cr 0 & 1 \cr } } \right]$$ then by the principle of mathematical induction, prove that $${A^n} = \left[ {\matrix{ 1 & {2n} \cr 0 & 1 \cr } } \right]$$

Answer

.

Explanation

Let given statement is denoted by P(n)

So, $$P(n):{A^n} = \left[ {\matrix{ 1 & {2n} \cr 0 & 1 \cr } } \right]$$ ..... (i)

Putting n = 1

$$P(1):{A^1} = \left[ {\matrix{ 1 & {2.1} \cr 0 & 1 \cr } } \right] = \left[ {\matrix{ 1 & 2 \cr 0 & 1 \cr } } \right] = A$$ (given)

$$\therefore$$ P(n) is true for n = 1

Let P(n) is true for n = k

i.e. $$P(k):{A^k} = \left[ {\matrix{ 1 & k \cr 0 & 1 \cr } } \right]$$ ..... (ii)

Now, we prove for n = k + 1 i.e.

$$P(k + 1):{A^{k + 1}} = \left[ {\matrix{ 1 & {2(k + 1)} \cr 0 & 1 \cr } } \right]$$ ..... (iii)

$$\therefore$$ $${A^{k + 1}} = {A^k} \times {A^1} = \left[ {\matrix{ 1 & {2k} \cr 0 & 1 \cr } } \right] \times \left[ {\matrix{ 1 & 2 \cr 0 & 1 \cr } } \right]$$ (from (ii))

$$ = \left[ {\matrix{ {1 \times 1 + 2k \times 0} & {1 \times 2 + 2k \times 1} \cr {0 \times 1 + 1 \times 0} & {0 \times 2 + 1 \times 1} \cr } } \right]$$

$$ = \left[ {\matrix{ 1 & {2 + 2k} \cr 0 & 1 \cr } } \right] = \left[ {\matrix{ 1 & {2(k + 1)} \cr 0 & 1 \cr } } \right]$$

$$\therefore$$ $${A^{k + 1}} = \left[ {\matrix{ 1 & {2(k + 1)} \cr 0 & 1 \cr } } \right]$$

$$\therefore$$ P(n) is true for n = k + 1 $$\therefore$$ P(n) is true for all n $$\in$$ N

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